Answer and Step-by-step explanation:
This is a complete question
Trials in an experiment with a polygraph include 97 results that include 23 cases of wrong results and 74 cases of correct results. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80% of the time. Identify the nullhypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
The computation is shown below:
The null and alternative hypothesis is
[tex]H_0 : p = 0.80[/tex]
[tex]Ha : p < 0.80[/tex]
[tex]\hat p = \frac{x}{ n} \\\\= \frac{74}{97}[/tex]
= 0.7629
Now Test statistic = z
[tex]= \hat p - P0 / [\sqrtP0 \times (1 - P0 ) / n][/tex]
[tex]= 0.7629 - 0.80 / [\sqrt(0.80 \times 0.20) / 97][/tex]
= -0.91
Now
P-value = 0.1804
[tex]\alpha = 0.01[/tex]
[tex]P-value > \alpha[/tex]
So, it is Fail to reject the null hypothesis.
There is ample evidence to demonstrate that less than 80 percent of the time reports that these polygraph findings are accurate.
Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6
Answer:
[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]
Step-by-step explanation:
Given that:
[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]
recall that:
cos (A-B) = cos AcosB + sin A sin B
∴
[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]
[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]
[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]
[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]
[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]
[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]
[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]
[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]
[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]
10) How many possible outfit combinations come from six shirts, three
slacks, and five ties? *
A 15
B 18
C 30
D 90
Answer:
The answer is D)90
Hope I helped
Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?
A)Right
B)Obtuse
C)Can't be determined
D) Acute
Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?
A)0.33 feet
B)3.75 feet
C)3 feet
D)5 feet
Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?
A)Acute
B)Right
C)Can't be determined
D)Obtuse
Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?
A)21.34 ft.
B)21.93 ft.
C)27.73 ft.
D)19.21 ft.
Answer:
Question 1 = D) Acute
Question 2 = C)3 feet
Question 3 = D) Obtuse
Question 4 = C)27.73 ft.
Step-by-step explanation:
Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?
In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem
Where:
If a² + b² = c² = Right angle triangle
If a² +b² > c² = Acute triangle.
If a² +b² < c² = Obtuse triangle.
It is important to note that the length ‘‘c′′ is always the longest.
Therefore, for the above question, we have lengths
5 = a, 6 = b and c = 7
a² + b² = c²
5² + 6² = 7²
25 + 36 = 49
61 = 49
61 ≠ 49, Hence 61 > 49
Therefore, this is an Acute Triangle
Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?
This is question that deals with proportion.
The formula to solve for this:
Height of the statue/ Length of the shadow of the person = Height of the person/ Length of the shadow of the person
Height of the statue = 15 feet
Length of the shadow of the person = 20 feet
Height of the person = unknown
Length of the shadow of the person = 4
15/ 20 = Height of the person/4
Cross Multiply
15 × 4 = 20 × Height of the person
Height of the person = 15 × 4/20
= 60/20
Height of the person = 3 feet
Therefore, the person is 3 feet tall.
Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?
In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem
Where:
If a² + b² = c² = Right angle triangle
If a² +b² > c² = Acute triangle.
If a² +b² < c² = Obtuse triangle.
It is important to note that the length ‘‘c′′ is always the longest.
Therefore, for the above question, we have lengths 17, 12, 9
9 = a, 12 = b and c = 17
a² + b² = c²
9² + 12² = 17²
81 + 144 = 289
225 = 289
225 ≠ 289
225 < 289
Hence, This is an Obtuse Triangle.
Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?
To calculate how far apart the two friends are we use the formula
Distance = √ ( Length² + Breadth²)
We are given dimensions: 12ft by 25ft
Length = 12ft
Breadth = 25ft
Distance = √(12ft)² + (25ft)²
Distance = √144ft²+ 625ft²
Distance = √769ft²
Distance = 27.730849248ft
Approximately ≈27.73ft
Therefore, the friends are 27.73ft apart.
In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal places.)
a. x=5
b. x <= 5
c. x>=6
Answer:
[tex]\mathbf{P(X=5) =0.0888}[/tex]
P(x ≤ 5 ) = 0.9707
P ( x ≥ 6) = 0.0293
Step-by-step explanation:
The probability of a binomial mass distribution can be expressed with the formula:
[tex]\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]
[tex]\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]
where;
n = 8 and π = 0.36
For x = 5
The probability [tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}[/tex]
[tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}[/tex]
[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}[/tex]
[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}[/tex]
[tex]\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}[/tex]
[tex]\mathtt{P(X=5) =0.0887645}[/tex]
[tex]\mathbf{P(X=5) =0.0888}[/tex] to 4 decimal places
b. x ≤ 5
The probability of P ( x ≤ 5)[tex]\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})[/tex]
[tex]{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +[/tex][tex]\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )[/tex]
P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888
P(x ≤ 5 ) = 0.9707
c. x ≥ 6
The probability of P ( x ≥ 6) = 1 - P( x ≤ 5 )
P ( x ≥ 6) = 1 - 0.9707
P ( x ≥ 6) = 0.0293
The age of some lecturers are 42,54,50,54,50,42,46,46,48 and 48 calculate the mean age and standard deviation
Answer:
Mean age: 48
Standard deviation: 4
Step-by-step explanation:
a) Mean
The formula for Mean = Sum of terms/ Number of terms
Number of terms
= 42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48+ 48/ 10
= 480/10
= 48
The mean age is 48
b) Standard deviation
The formula for Standard deviation =
√(x - Mean)²/n
Where n = number of terms
Standard deviation =
√[(42 - 48)² + (54 - 48)² + (50 - 48)² +(54 - 48)² + (50 - 48)² +(42 - 48)² + (46 - 48)² + (46 - 48)² + (48 - 48)² + (48 - 48)² / 10]
= √-6² + 6² + 2² + 6² + 2² + -6² + -2² + -2² + 0² + 0²/10
=√36 + 36 + 4 + 36 + 4 + 36 + 4 + 4 + 0 + 0/ 10
=√160/10
= √16
= 4
The standard deviation of the ages is 4
Find the principal invested if $495 interest was earned in 3 years at an interest rate of 6%.
Answer: $2750
Step-by-step explanation:
Formula to calculate interest : I = Prt , where P = Principal amount , r = rate of interest ( in decimal) , t= time.
Given: I= $495
t= 3 years
r= 6% = 0.06
Then, according to the above formula:
[tex]495 = P (0.06\times3)\\\\\Rightarrow\ P=\dfrac{495}{0.18}\\\\\Rightarrow\ P=2750[/tex]
Hence, the principal invested = $2750
Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.
Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]
$a^{2}=5+2 \sqrt{6}$
$a^{3}=11 \sqrt{2}+9 \sqrt{3}$
The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.
Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$
so fits with the other answers.
Answer:
[tex]y^3 -6y-6[/tex]
The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:
Complete Question
On the uploaded image is a similar question that will explain the given question
Answer:
The value of k is [tex]k = 214285.7[/tex]
The percentage of the oil that will be cleaned is [tex]x = 80.77\%[/tex]
Step-by-step explanation:
From the question we are told that
The cost of cleaning up the spillage is [tex]C = \frac{ k x }{100 - x }[/tex] [tex]x \le x \le 100[/tex]
The cost of cleaning x = 70% of the oil is [tex]C = \$500,000[/tex]
Now at [tex]C = \$500,000[/tex] we have
[tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]k = 214285.7[/tex]
Now When [tex]C = \$900,000[/tex]
[tex]x = 80.77\%[/tex]
A senior accounting major at Midsouth State University has job offers from four CPA firms. To explore the offers further, she asked a sample of recent trainees how many months each worked for the firm before receiving a raise in salary. The sample information is submitted to Minitab with the following results:
Analysis of Variance
Source df SS MS F P
Factor 3 28.17 9.39 5.37 0.010
Error 15 26.26 1.75
Total 18 54.43
A) Reject H0 if F >
B) For the 0.05 level of significance, is there a difference in the mean difference in the mean number of months before a raise was anted among the four CPA firms?
Answer:
A) Reject H0 if F > 5.417
B) we fail to reject the null hypothesis and conclude that we do not have sufficient evidence at 0.05 level of significance to support the claim that there is a difference in the mean number of months before a raise was granted among the four CPA firms
Step-by-step explanation:
A) From the table, we can see that we have df1 = 3 and df2 = 15. And we are given a significance level of α = 0.01
We are also given f-value of 1.75
Thus,from the f-distribution table attached at significance level of α = 0.01 and df1 = 3 and df2 = 15, we have;
F-critical = 5.417
Normally, we reject H0 if F > 5.417
But in this case, F is 1.75 < 5.417 and so we conclude that we do not reject H0 at the 0.01 level of significance
B) for 0.05 level of significance, df1 = 3 and df2 = 15, from the 2nd table attached, we have;
F-critical = 3.2874
Again the f-value is less than this critical one.
Thus, we fail to reject the null hypothesis and conclude that we do not have sufficient evidence at 0.05 level of significance to support the claim that there is a difference in the mean number of months before a raise was granted among the four CPA firms
one third multiplied by the sum of a and b
Answer:
1/3(a+b)
hope it helps :>
Type the missing number in this sequence:
1,
4,
,64, 256,
1,024
Answer:
16
Step-by-step explanation:
The sequence is 1, 4,...,64, 256, 1024
Notice that:
● 1 = 2^0
● 4 = 2^2
● 64 = 2^6
● 256 = 2^8
● 1024 = 2^10
Notice that we add 2 each time to the exponent so the missing number is:
● 2^(2+2) = 2^4 = 16
Transform the given parametric equations into rectangular form. Then identify the conic.
Answer:
Solution : Option B
Step-by-Step Explanation:
We have the following system of equations at hand here.
{ x = 5 cot(t), y = - 3csc(t) + 4 }
Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,
x = 5 cot(t) ⇒ x - 5 = cot(t),
y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)
Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations as well. --- Step #2
( y - 4 / - 3 )² = (csc(t))²
- ( x - 5 / 1 )² = (cot(t))²
___________________
(y - 4)² / 9 - x² / 25 = 1
And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.
Please help ! I’ll mark you as brainliest if correct.
Answer:
D = -87Dx = 174Dy = -435Dz = 0(x, y, z) = (-2, 5, 0)Step-by-step explanation:
The determinant of the coefficient matrix is ...
[tex]D=\left|\begin{array}{ccc}2&5&3\\4&-1&-4\\-5&-2&6\end{array}\right|\\\\=2(-1)(6)+5(-4)(-5)+3(4)(-2)-2(-4)(-2)-5(4)(6)-3(-1)(-5)\\\\=-12+100-24-16-120-15=\boxed{-87}[/tex]
The other determinants are found in similar fashion after substituting the constants on the right for each of the above matrix columns, in turn.
Those determinants are ...
[tex]D_x=\left|\begin{array}{ccc}21&5&3\\-13&-1&-4\\0&-2&6\end{array}\right|=174[/tex]
[tex]D_y=\left|\begin{array}{ccc}2&21&3\\4&-13&-4\\-5&0&6\end{array}\right|=-435[/tex]
[tex]D_z=\left|\begin{array}{ccc}2&5&21\\4&-1&-13\\-5&-2&0\end{array}\right|=0[/tex]
The solutions are ...
x = 174/-87 = -2
y = -435/-87 = 5
z = 0
That is, (x, y, z) = (-2, 5, 0).
You are ordering two pizzas. A pizza can be small, medium, large, or extra large, with any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8). How many possibilities are there for your two pizzas
Answer:
1048576
Step-by-step explanation:
Given the following :
Pizza order :
Size = small, medium, large, or extra large = 4 possible sizes
Toppings = any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8).
Combination of Toppings = 2^8
Four different sizes of pizza = 4
Number of possibilities in ordering for a single pizza :
(4 * 2^8) = 4 * 256 = 1024
Number of possibilities in ordering two pizzas :
(4 * 2^8)^2
(2^2 * 2^8)^2
From indices :
[2^(2+8)]^2
[2^(10)]^2
2^(10*2)
2^20
= 1048576
find the perimeter of a square of sides 10.5cm
Answer:
Perimeter = 42 cm
Step-by-step explanation:
A square has all equal sides so you would just add 10.5 + 10.5 + 10.5 + 10.5 to get 42 cm.
Answer:
42 cm
Step-by-step explanation:
Side of square = 10.5 cm (given)
Perimeter of square = Side X 4
= 10.5 X 4
= 42 cm
HOPE THIS HELPED YOU !
:)
Gail paid a total of $12,000 for stock that was $6 per share. If she sold all her shares for $18,000, how much profit on each share did she make?
A
$9
B
$3
С.
S2000
D
$6.000
Answer:
$3
Step-by-step explanation:
Given
Total Cost Price: $12,000
Unit Cost Price= $6
Total Selling Price = $18,000
Required
Determine the profit on each share
First, we need to determine the units of share bought;
Units = Total cost price / Unit Cost Price
[tex]Units = \frac{\$12000}{\$6}[/tex]
[tex]Units = 2000[/tex]
Next is to determine the selling price of each share; This is calculated as follows;
Unit Selling Price = Total Selling Price / Units Sold
[tex]Unit\ Selling\ Price = \frac{\$18000}{\$2000}[/tex]
[tex]Unit\ Selling\ Price = \$9[/tex]
The profit is the difference between the unit cost price and unit selling price
[tex]Profit = Unit\ Selling\ Price - Unit\ Cost\ Price[/tex]
[tex]Profit = \$9 - \$6[/tex]
[tex]Profit = \$3[/tex]
Three out of every ten dentists recommend a certain brand of fluoride toothpaste. Which assignment of random digits would be used to simulate the random sampling of dentists who prefer this fluoride toothpaste?
Answer:
eddfdgdccggģdffcdrrfxddxcvgfx
logx-log(x-l)^2=2log(x-1)
Answer:
x = 1.00995066776
x = 2.52925492433
Step-by-step explanation:
This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...
[tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]
The attached graph shows zeros at
x = 1.00995066776 and 2.52925492433
_____
Comment on the equation
Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068
Comment on the answer refinement
We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H0: μ = 100 versus H1: μ ≠ 100 with a sample of n = 9 specimens.
A. If the acceptance region is defined as 98.5 le x- 101.5, find the type I error probability alpha.
B. Find beta for the case where the true mean heat evolved is 103.
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
Answer:
A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]
B. β = 0.0122
C. β = 0.0000
Step-by-step explanation:
Given that:
Mean = 100
standard deviation = 2
sample size = 9
The null and the alternative hypothesis can be computed as follows:
[tex]\mathtt{H_o: \mu = 100}[/tex]
[tex]\mathtt{H_1: \mu \neq 100}[/tex]
A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .
Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]
∴
[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]
[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]
when [tex]\mu = 100[/tex]
[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]
[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]
[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]
From the standard normal distribution tables
[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]
[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]
[tex]\mathbf{\alpha = 0.0244 }[/tex]
Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]
B. Find beta for the case where the true mean heat evolved is 103.
The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]
Thus;
β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )
∴
[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]
Given that [tex]\mu = 103[/tex]
[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]
[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]
[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]
From standard normal distribution table
β = 0.0122 - 0.0000
β = 0.0122
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]
Given that [tex]\mu = 105[/tex]
[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]
[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]
[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]
From standard normal distribution table
β = 0.0000 - 0.0000
β = 0.0000
The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.
Please help me solve for the median !!!
Answer:
50.93
Step-by-step explanation:
Add up the frequencies:
2 + 5 + 14 + 15 + 21 + 18 + 15 + 9 + 2 = 101
Divide by 2: 101/2 = 50.5
So the median is the 51st number, with 50 below and 50 above.
Add up the frequencies until you find the interval that contains the 51st number.
2 + 5 + 14 + 15 = 36
2 + 5 + 14 + 15 + 21 = 57
So the median is in the group 49.5 − 51.5. To estimate the median, we use interpolation. Find the slope of the line from (36, 49.5) to (57, 51.5).
m = (51.5 − 49.5) / (57 − 36)
m = 2/21
So at x = 51:
2/21 = (y − 49.5) / (51 − 36)
y = 50.93
which of the following not between -10 and -8
-17/2
-7
-9
-8.5
Answer:
-7Step-by-step explanation:
This is best read on the number line.
Look at the picture.
[tex]-\dfrac{17}{2}=-8\dfrac{1}{2}=-8.5[/tex]
Use the two highlighted points to find the
equation of a trend line in slope-intercept
form.
Answer: y=(4/3)x+2/3
Step-by-step explanation:
Slope-intercept form is expressed as y=mx+b
First, find the slope (m):
m= rise/run or vertical/horizontal or y/x (found between the highlighted points)
m = 4/3
Second, find b:
Use one of the highlighted points for (x, y)
2=4/3(1)+b
6/3=4/3+b
2/3=b
b=2/3
Plug it into the equation:
You get y=(4/3)x+2/3 :)
At a local high school, the student population is growing at 12% a year. If the original population was 242 students, how long will it take the population to reach 300 students? Round to the nearest tenth of a year.
Answer: 2 years
Step-by-step explanation:
The exponential growth function is given by :-
[tex]y=A(1+r)^x[/tex] (i)
, where A = initial value , r = rate of growth and x= time period.
As per given ,
A= 242
r= 12% = 0.12
To find : t when y= 300.
Put all the values in (i)
[tex]300=242(1+0.12)^x\\\\\Rightarrow\ \dfrac{300}{242}=(1.12)^x\\\\\Rightarrow\ 1.23967=(1.12)^x[/tex]
Taking log on both sides , we get
[tex]\log (1.2396) = t \log (1.12)\\\\\Rightarrow\ 0.09328=t(0.049218)\\\\\Rightarrow t=\dfrac{0.09328}{0.049218}=\approx2[/tex]
hence, it will take 2 years.
Write in words how we would say the following
3 square
Answer:
Three to the second power
Step-by-step explanation:
Hey there!
3 square
Can be written as the following,
Three to the second power
Hope this helps :)
Use Lagrange multipliers to minimize the function subject to the following two constraints. Assume that x, y, and z are nonnegative. Question 18 options: a) 192 b) 384 c) 576 d) 128 e) 64
Complete Question
The complete question is shown on the first uploaded image
Answer:
Option C is the correct option
Step-by-step explanation:
From the question we are told that
The equation is [tex]f (x, y , z ) = x^2 +y^2 + z^2[/tex]
The constraint is [tex]P(x, y , z) = x + y + z - 24 = 0[/tex]
Now using Lagrange multipliers we have that
[tex]\lambda = \frac{ \delta f }{ \delta x } = 2 x[/tex]
[tex]\lambda = \frac{ \delta f }{ \delta y } = y[/tex]
[tex]\lambda = \frac{ \delta f }{ \delta z } = 2 z[/tex]
=> [tex]x = \frac{ \lambda }{2}[/tex]
[tex]y = \frac{ \lambda }{2}[/tex]
[tex]z = \frac{ \lambda }{2}[/tex]
From the constraint we have
[tex]\frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{2} = 24[/tex]
=> [tex]\frac{3 \lambda }{2} = 24[/tex]
=> [tex]\lambda = 16[/tex]
substituting for x, y, z
=> x = 8
=> y = 8
=> z = 8
Hence
[tex]f (8, 8 , 8 ) = 8^2 +8^2 + 8^2[/tex]
[tex]f (8, 8 , 8 ) = 192[/tex]
Hey market sales six cans of food for every seven boxes of food the market sold a total of 26 cans and boxes today how many of each kind did the market sale
Answer:
It sold 14 cans boxes of food and 12 cans of food.
Step-by-step explanation:
The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.
Let the no. of sets of food boxes be x.
According to the question,
6x+7x=26
13x=26
x=26/13
x=2
No. of food cans =6x=6×2=12 cans
No. of food boxes=7x=7×2=14 boxes
Please mark brainliest ,if it is truly the best ! Thank you!
5x+4(-x-2)=-5x+2(x-1)+12
Answer:
x=9/2
Step-by-step explanation:
Let's solve your equation step-by-step.
5x+4(−x−2)=−5x+2(x−1)+12
Step 1: Simplify both sides of the equation.
5x+4(−x−2)=−5x+2(x−1)+12
5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)
5x+−4x+−8=−5x+2x+−2+12
(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)
x+−8=−3x+10
x−8=−3x+10
Step 2: Add 3x to both sides.
x−8+3x=−3x+10+3x
4x−8=10
Step 3: Add 8 to both sides.
4x−8+8=10+8
4x=18
Step 4: Divide both sides by 4.
4x/4=18/4
x=9/2
A regular polygon inscribed in a circle can be used to derive the formula for the area of a circle. The polygon area can be expressed in terms of the area of a triangle. Let s be the side length of the polygon, let r be the hypotenuse of the right triangle, let h be the height of the triangle, and let n be the number of sides of the regular polygon. polygon area = n(12sh) Which statement is true? As h increases, s approaches r so that rh approaches r². As r increases, h approaches r so that rh approaches r². As s increases, h approaches r so that rh approaches r². As n increases, h approaches r so that rh approaches r².
Answer:
Option (D)
Step-by-step explanation:
Formula to get the area of a regular polygon in a circle will be,
Area = [tex]n[\frac{1}{2}\times (\text{Base})\times (\text{Height})][/tex]
= [tex]n[\frac{1}{2}\times (\text{s})\times (\text{h})][/tex]
Here 'n' is the number of sides.
If n increases, h approaches r so that 'rh' approaches r².
In other words, if the number of sides of the polygon gets increased, area of the polygon approaches the area of the circle.
Therefore, Option (4) will be the answer.
In this exercise it is necessary to have knowledge about polygons, so we have to:
Letter D
Then using the formula for the area of a regular polygon we find that:
[tex]A=n(1/2*B*H)\\=n(1/2*S*H)[/tex]
So from this way we were not able to identify the option that best corresponds to this alternative.
See more about polygons at brainly.com/question/17756657
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a standard deviation of ounce. You randomly select cans and carefully measure the contents. The sample mean of the cans is ounces. Does the machine need to be reset? Explain your reasoning. ▼ Yes No , it is ▼ very unlikely likely that you would have randomly sampled cans with a mean equal to ounces, because it ▼ lies does not lie within the range of a usual event, namely within ▼ 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means.
Complete question is;
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.
(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.
Answer:
Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
Step-by-step explanation:
We are given;
Mean: μ = 128
Standard deviation; σ = 0.2
n = 35
Now, formula for standard error of mean is given as;
se = σ/√n
se = 0.2/√35
se = 0.0338
Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;
μ ± 2se = 128 ± 0.0338
This gives; 127.9662, 128.0338
So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
cooks are needed to prepare for a large party. Each cook can bake either 5 Large cakes or 14 small cakes per hour . The kitchen is available for 3 hours and 29 large cakes and 260 cakes need to be baked . How many cooks are required to bake the required number of cakes during the time the kitchen is available?
it was all about equating some values
to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.
Let's determine the number of cooks required to bake the required number of cakes during the available time.
We have the following information:
- Each cook can bake either 5 large cakes or 14 small cakes per hour.
- The kitchen is available for 3 hours.
- We need to bake 29 large cakes and 260 cakes in total.
First, let's calculate the number of large cakes that can be baked by one cook in 3 hours:
1 cook can bake 5 large cakes/hour × 3 hours = 15 large cakes.
Next, let's calculate the number of small cakes that can be baked by one cook in 3 hours:
1 cook can bake 14 small cakes/hour × 3 hours = 42 small cakes.
Now, let's calculate the number of large cakes that can be baked by all the cooks in 3 hours:
Total number of large cakes = Number of cooks × Large cakes per cook per 3 hours
We need to bake 29 large cakes, so:
29 = Number of cooks × 15
Number of cooks = 29 / 15 ≈ 1.93
Since we can't have a fraction of a cook, we need to round up to the nearest whole number. Therefore, we need at least 2 cooks to bake the required number of large cakes.
Similarly, let's calculate the number of small cakes that can be baked by all the cooks in 3 hours:
Total number of small cakes = Number of cooks × Small cakes per cook per 3 hours
We need to bake 260 small cakes, so:
260 = Number of cooks × 42
Number of cooks = 260 / 42 ≈ 6.19
Again, rounding up to the nearest whole number, we need at least 7 cooks to bake the required number of small cakes.
Since we need to satisfy both requirements for large and small cakes, we choose the larger number of cooks required, which is 7 cooks.
Therefore, to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.
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