Answer:
s is [tex]s =15.3[/tex]
Step-by-step explanation:
From the question we are told that
The data is
Generally the mean is evaluated as
[tex]\= x = \frac{84+ 85+ 83+ 63+ 61+100+ 98}{7}[/tex]
[tex]\= x = 82[/tex]
The standard deviation is mathematically represented as
[tex]s = \frac{\sum (x- \= x)^2}{n-1}[/tex]
[tex]s =\sqrt{ \frac{ (84-82)^2+ (85-82)^2 +( 83-82 )^2 +(63-82) ^2 + ( 61-82)^2,+ (100-82)^2 + (98-82)^2}{7-1}}[/tex]
[tex]s =15.3[/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!
Records indicate that x years after 2008, the average property tax on a three bedroom home in a certain community was T(x) =20x^2+40x+600 dollars.
Required:
a. At what rate was the property tax increasing with respect to time in 2008?
b. By how much did the tax change between the years 2008 and 2012?
Answer:
a) 40 dollars
b) 480 dollars
Step-by-step explanation:
Given the average property tax on a three bedroom home in a certain community modelled by the equation T(x) =20x²+40x+600, the rate at which the property tax is increasing with respect to time in 2008 can be derived by solving for the function T'(x) at x=0
T'(x) = 2(20)x¹ + 40x° + 0
T'(x) = 40x+40
At x = 0,
T'(0) = 40(0)+40
T'(0) = 40
Hence the property tax was increasing at a rate of 40dollars with respect to the initial year (2008).
b) There are 4 years between 2008 and 2012. To know how much that the tax change between the years 2008 and 2012, we will find T(4) - T(0)
Given T(x) =20x²+40x+600
T(4) =20(4)²+40(4)+600
T(4) = 320+160+600
T(4) = 1080 dollars
Also T(0) =20(0)²+40(0)+600
T(0) = 0+0+600
T(0)= 600 dollars
T(4) - T(0) = 1080 - 600
T(4) - T(0) = 480 dollars
Hence, the tax has changed by $480 between 2008 and 2012
Four couples are at a party. Four of the eight people are randomly selected to win a prize. No person can win more than one prize. What is the probability that both of the members of at least one couple win prizes? Express your answer as common fraction.
Answer:
27/35
Step-by-step explanation:
We use combination to solve for this
C(n, r), =nCr = n!/r!(n - r)!
From the question, we are told that:
Four couples are at a party. Four of the eight people are randomly selected to win a prize.
Four couples = 8 people.
= 8C4 = 8!/4! (8 - 4)!
= 70
No person can win more than one prize. ( No person can win more than one prize of the 4 people selected)
This can happen in 4 ways
[4C1 × 3C2 ] × 4=
[4!/1! ×( 4 - 1)!] × [3!/2! ×(3-2)!] × 4 ways
= 4 × 3 × 4 ways
= 48
The probability that both of the members of at least one couple win prizes
48 + 4C2/ 8C4
4C2 = 4!/2!(4 - 2) !
= 6
8C4 = 8C4 = 8!/4! (8 - 4)!
= 70
48 + 6/ 70
= 54/70
= 27/35
Therefore, the probability that both of the members of at least one couple win prizes is 27/35.
The probability that both of the members of at least one couple win prizes is 27/35 and this can be determined by using the given data.
Given :
Four couples are at a party. Four of the eight people are randomly selected to win a prize. No person can win more than one prize.The following steps can be used in order to determine the probability that both of the members of at least one couple win prizes:
Step 1 - The concept of probability is used in order to determine the probability that both of the members of at least one couple win prizes.
Step 2 - According to the given data, the total number of people is 8.
Step 3 - So, the probability that both of the members of at least one couple win prizes is:
[tex]\rm P =\dfrac{ \;^4C_1\times \;^3C_2\times 4 + \;^4C_2}{\;^8C_4}[/tex]
Step 4 - Simplify the above expression.
[tex]\rm P =\dfrac{48+ 6}{70}[/tex]
[tex]\rm P = \dfrac{27}{35}[/tex]
So, the probability that both of the members of at least one couple win prizes is 27/35.
For more information, refer to the link given below:
https://brainly.com/question/795909
In the Cash Now lottery game there are 8 finalists who submitted entry tickets on time. From these 8 tickets, three grand prize winners will be drawn. The first prize is one million dollars, the second prize is one hundred thousand dollars, and the third prize is ten thousand dollars. Determine the total number of different ways in which the winners can be drawn. (Assume that the tickets are not replaced after they are drawn.)
Answer:
The number of ways is [tex]\left n} \atop {}} \right. P_r = 336[/tex]
Step-by-step explanation:
From the question we are told that
The number of tickets are [tex]n = 8[/tex]
The number of finalist are [tex]r =3[/tex]
Generally the number of way by which this winners can be drawn and arrange in the order of [tex]1^{st} , \ 2nd , \ 3rd[/tex] is mathematically represented as
[tex]\left n} \atop {}} \right. P_r = \frac{n\ !}{(n-r) !}[/tex]
substituting values
[tex]\left n} \atop {}} \right. P_r = \frac{ 8!}{(8-3) !}[/tex]
[tex]\left n} \atop {}} \right. P_r = \frac{ 8* 7*6*5*4*3*2*1}{ 5*4*3*2*1}[/tex]
[tex]\left n} \atop {}} \right. P_r = 336[/tex]
Given log32≈0.631 and log37≈1.771, what is log314
Answer:
the log to the base 3 of 14 is 2.402
Step-by-step explanation:
You must find a way to indicate that 3 is the base; you cannot run this '3' together with 2, 7 or 14.
Example:
log to the base 3 of 2 = 0.631
log to the base 3 of 7 = 1.771
Note that 2 times 7 is 14. Thus, to obtain the log to the base 3 of 14, we must ADD the two logs shown above:
0.631
+1.771
----------
2.402
Thus, the log to the base 3 of 14 is 2.402.
Check: Does 3^2.402 = 14? YES
Determine the volume of a sphere with a diameter of 70 mm. Question 13 options: A) 21,714.7 mm3 B) 3,216.9 mm3 C) 100,024 mm3 D) 179,594.4 mm3
Answer:
The answer is option D
Step-by-step explanation:
Volume of a sphere is given by
[tex]V = \frac{4}{3} \pi {r}^{3} [/tex]
where r is the radius
From the question to calculate the radius we use the formula
radius = diameter / 2
diameter =70mm
radius = 70/2 = 35 mm
So the volume of the sphere is
[tex]V = \frac{4}{3} \pi \times {35}^{3} [/tex]
[tex]V = \frac{171500\pi}{3} [/tex]
We have the final answer as
Volume = 179,594.4 mm³Hope this helps you
What is nine thousandths as a decimal
Answer:
Nine thousandths = 0.009
Step-by-step explanation:
thousandths = 1/1000 = 0.001
nine thousandths = 9/1000 = 0.009
Answer:
.009
Step-by-step explanation:
9 thousandths as a decimal is 9/1000. Which is the same 0.009
determine x in the following equation 2x - 4 = 10
Answer:
7
Step-by-step explanation:
10+4 = 14
14/2 = 7
x = 7
If SSR is 2592 and SSE is 608, then A. the standard error would be large. B. the coefficient of determination is .23. C. the slope is likely to be insignificant. D. the coefficient of determination is .81.
Answer:
D. the coefficient of determination is .81.
Step-by-step explanation:
SST = SSE + SSR
where
SST is the summation of square total
SSE is the summation of squared error estimate = 608
SSR is the summation of square of residual = 2593
with these in mind we put the values into the formula
= 2592 + 608
=3200
Coefficient of determination = SSR/SST
= 2592/3200
= 0.81
Therefore option D is the correct answer to the question.
There are 9 students at the math club picnic. If 3 students are drinking punch and 6 are drinking lemonade, what fraction are drinking lemonade
A sample of 81 observations is taken from a normal population with a standard deviation of 5. The sample mean is 40. Determine the 95% confidence interval for the population mean.
Answer:
38.911≤p≤41.089
Step-by-step explanation:
The formula for calculating confidence interval for a population mean us as shown below;
CI = xbar ± Z×S/√N where;
xbar is the sample mean = 40
Z is the z score at 95% confidence interval = 1.96
S is the standard deviation = 5
N is the sample size = 81
Substituting this parameters in the formula we have;
CI = 40±1.96×5/√81
CI = 40±(1.96×5/9)
CI = 40±(1.96×0.556)
CI = 40±1.089
CI = (40-1.089, 40+1.089)
CI = (38.911, 41.089)
The 95% confidence interval for the population mean is 38.911≤p≤41.089
Answer:
38.9 ≤ U ≤ 41.1
Step-by-step explanation:
Mean, m = 40; standard deviation, α = 5; Confidence limit, U = 95% or 0.95
N = 81
The standard error, α(m) = α/√(N) = 5/√81 =5/9
Using table: 0.95 = 0.0379
Z(0.95) = 2 - 0.0379 = 1.9621 or 1.96
Hence, confidence interval = { m - 1.96(α/√N) ≤ U ≤ m +1.96(α/√N)}
But, 1.96(α/√N) = 1.96 X 5/9 = 1.96 X 0.56 = 1.1
(40 - 1.1 ≤ U ≤ 40 + 1.1)
∴ the confidence interval = 38.9 ≤ U ≤ 41.1
what is the number if 4 is subtracted from the sum of one fourth of 5 times of 8 and 10
Answer:
Step-by-step explanation:
Lets, turn this into words and use order of operations, First, we look for multiplication and division.
the sum of one fourth of 5 times of 8 and 10 gets you 1/4(5*8) + 10 = 20
what is the number if 4 is subtracted from the sum
20 - 4 = 16
Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours. As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.
A) What can you say about the shape of the distribution of the sample mean?
B) What is the standard error of the distribution of the sample mean?
C) What proportion of the samples will have a mean useful life of more than 36 hours?
D) What proportion of the sample will have a mean useful life greater than 34.5 hours?
E) What proportion of the sample will have a mean useful life between 34.5 and 36.0 hours?
Answer:
(A) The shape of the distribution of the sample mean is bell-shaped.
(B) The standard error of the distribution of the sample mean is 1.1.
(C) The proportion of the samples that have a mean useful life of more than 36 hours is 0.1814.
(D) The proportion of the sample that has a mean useful life greater than 34.5 hours is 0.6736.
(E) The proportion of the sample that has a mean useful life between 34.5 and 36.0 hours is 0.4922.
Step-by-step explanation:
We are given that Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours.
As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.
Let [tex]\bar X[/tex] = sample mean life of these batteries
(A) The shape of the distribution of the sample mean will be bell-shaped because the sample mean also follows the normal distribution as it is taken from the population data only.
(B) The standard error of the distribution of the sample mean is given by;
Standard error = [tex]\frac{\sigma}{\sqrt{n} }[/tex]
Here, [tex]\sigma[/tex] = standard deviation = 5.5 hours
n = sample of batteries = 25
So, the standard error = [tex]\frac{5.5}{\sqrt{25} }[/tex] = 1.1.
(C) The z-score probability distribution for the sample mean is given by;
Z = [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean life of battery = 35.0 hours
[tex]\sigma[/tex] = standard deviation = 5.5 hours
n = sample of batteries = 25
Now, the proportion of the samples that will have a mean useful life of more than 36 hours is given by = P([tex]\bar X[/tex] > 36 hours)
P([tex]\bar X[/tex] > 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > 0.91) = 1 - P(Z [tex]\leq[/tex] 0.91)
= 1 - 0.8186 = 0.1814
(D) The proportion of the samples that will have a mean useful life of more than 34.5 hours is given by = P([tex]\bar X[/tex] > 34.5 hours)
P([tex]\bar X[/tex] > 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > -0.45) = P(Z [tex]\leq[/tex] 0.45)
= 0.6736
(E) The proportion of the samples that will have a mean useful life between 34.5 and 36.0 hours is given by = P(34.5 hrs < [tex]\bar X[/tex] > 36 hrs)
P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = P([tex]\bar X[/tex] < 36 hrs) - P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hrs)
P([tex]\bar X[/tex] < 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z < 0.91) = 0.8186
P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.45) = 1 - P(Z [tex]\leq[/tex] 0.45)
= 1 - 0.6736 = 0.3264
Therefore, P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = 0.8186 - 0.3264 = 0.4922.
Help please!!! Thank you
Answer:
5/7
Step-by-step explanation:
There are a couple ways to solve this. One would be by finding the least common denominator for each one with 2/3, subtracting, and seeing what is left over. Another way is converting to decimals.
2/3=0.666666
————————-
7/8=0.875
8/9=0.88888
4/5=0.8
5/7=0.7143
They are all greater than 2/3 (0.6666666), but 5/7 is the closest, so would have the least waste.
There are $400$ pages in Sheila's favorite book. The average number of words per page in the book is $300$. If she types at an average rate of $40$ words per minute, how many hours will it take to type the $400$ pages of the book?
Answer:
50hours
Step-by-step explanation:
Given that there are 400 pages in Sheila's favorite book.
The average number of words per page in the book is 300
She types an average rate of 40words per minute.
So to type 400pages of the book
Total number of words in the pages = 400×300 = 120000 words
Typing rate : 40words ------- 1minute
120000 words ----------- x minutes
Hence we have 40 × X mins = 120000 × 1min
Make X the subject
40X = 120000minutes
X = 120000/40
X = 3000minutes
Since 60minutes = 1hour
3000minutes = 3000minutes/60
= 50hours
Hence it took her 50hours to type 400pages
Solution:
The total number of words in the book is 400 x 300. Sheila types at a rate of 40 words per minute, or 40 x 60 words per hour. The number of hours it takes her is equal to the number of words divided by her rate of typing, or 400x300/40x60 = 50 hours.
Height of a tree increases by 2.5 feet each growing season. Quadratic, linear or exponential?
Answer:
Linear
Step-by-step explanation:
Given
Height of a tree grows by 2.5 feet
Required
Determine the type of relationship
Take for instance, the height of the tree at year 1 is x
At year 2, it will be x + 2 * 1
At year 3, it will be x + 2 * 2
At year 4, it will be x + 2 * 3
Following same pattern
At year n, it will be x + 2 *(n - 1)
Hence, growth rate = x + 2(n -1)
From the list of given options, the correct answer is Linear because the derived formula above is an example of a linear equation
Study the table. Which best describes the function represented by the data in the table?
Answer:
linear with a common first difference of 2
Step-by-step explanation:
On the face of it, you can reject answers that ascribe a common ratio to a linear or quadratic function. (A common ratio is characteristic of an exponential function.)
You can also reject the answer that ascribes a common first difference to a quadratic function. (A quadratic function has a common second difference.)
After you reject the nonsense answers, there is only one remaining choice. It is also the correct one:
linear with a common first difference of 2
_____
The ratio of change in y to change in x is ...
(0 -(-2))/(-2 -(-3)) = 2
(4 -0)/(0 -(-2)) = 2
(12 -4)/(4 -0) = 2
That is, y increases by 2 when x increases by 1. The common first difference is 2.
Given the number of trials and the probability of success, determine the probability indicated: a. n = 15, p = 0.4, find P(4 successes) b. n = 12, p = 0.2, find P(2 failures) c. n = 20, p = 0.05, find P(at least 3 successes)
Answer:
A)0.126775 B)0.000004325376 C) 0.07548
Step-by-step explanation:
Given the following :
A.) a. n = 15, p = 0.4, find P(4 successes)
a = number of trials p=probability of success
P(4 successes) = P(x = 4)
USING:
nCx * p^x * (1-p)^(n-x)
15C4 * 0.4^4 * (1-0.4)^(15-4)
1365 * 0.0256 * 0.00362797056
= 0.126775
B)
b. n = 12, p = 0.2, find P(2 failures),
P(2 failures) = P(12 - 2) = p(10 success)
USING:
nCx * p^x * (1-p)^(n-x)
12C10 * 0.2^10 * (1-0.2)^(12-10)
66 * 0.0000001024 * 0.64
= 0.000004325376
C) n = 20, p = 0.05, find P(at least 3 successes)
P(X≥ 3) = p(3) + p(4) + p(5) +.... p(20)
To avoid complicated calculations, we can use the online binomial probability distribution calculator :
P(X≥ 3) = 0.07548
The mean weight of newborn infants at a community hospital is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. Does the sample data show a significant increase in the average birthrate at a 5% level of significance?
A. Fail to reject the null hypothesis and conclude the mean is 6.6 lb.
B. Reject the null hypothesis and conclude the mean is lower than 6.6 lb.
C. Reject the null hypothesis and conclude the mean is greater than 6.6 lb.
D. Cannot calculate because the population standard deviation is unknown
Answer:
The correct option is A
Step-by-step explanation:
From the question we are told that
The population is [tex]\mu = 6.6[/tex]
The level of significance is [tex]\alpha = 5\% = 0.05[/tex]
The sample data is 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds
The Null hypothesis is [tex]H_o : \mu = 6.6[/tex]
The Alternative hypothesis is [tex]H_a : \mu > 6.6[/tex]
The critical value of the level of significance obtained from the normal distribution table is
[tex]Z_{\alpha } = Z_{0.05 } = 1.645[/tex]
Generally the sample mean is mathematically evaluated as
[tex]\=x = \frac{\sum x_i }{n}[/tex]
substituting values
[tex]\=x = \frac{9.0 + 7.3 + 6.0+ 8.8+ 6.8+ 8.4+6.6 }{7}[/tex]
[tex]\=x = 7.5571[/tex]
The standard deviation is mathematically evaluated as
[tex]\sigma = \sqrt{\frac{\sum [ x - \= x ]}{n} }[/tex]
substituting values
[tex]\sigma = \sqrt{\frac{ [ 9.0-7.5571]^2 + [7.3 -7.5571]^2 + [6.0-7.5571]^2 + [8.8- 7.5571]^2 + [6.8- 7.5571]^2 + [8.4 - 7.5571]^2+ [6.6- 7.5571]^2 }{7} }[/tex][tex]\sigma = 1.1774[/tex]
Generally the test statistic is mathematically evaluated as
[tex]t = \frac{\= x - \mu } { \frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
[tex]t = \frac{7.5571 - 6.6 } { \frac{1.1774 }{\sqrt{7} } }[/tex]
[tex]t = 1.4274[/tex]
Looking at the value of t and [tex]Z_{\alpha }[/tex] we see that [tex]t < Z_{\alpha }[/tex] hence we fail to reject the null hypothesis
What this implies is that there is no sufficient evidence to state that the sample data show as significant increase in the average birth rate
The conclusion is that the mean is [tex]\mu = 6.6 \ lb[/tex]
Factor 4(20) + 84. 4(20 + 21) 4(21 + 20) 20(4 + 84) 20(4 + 4)
Answer:
[tex]\huge\boxed{4 ( 20 + 21)}[/tex]
Step-by-step explanation:
4(20) + 84
Resolve Parenthesis
80 + 84
Taking 4 common as both are the multiples of 4
4 ( 20 + 21)
Which equation does the graph of the systems of equations solve? (1 point) 2 linear graphs. They intersect at negative 1, 1
Answer:
3x +4 = -2x -1
Step-by-step explanation:
The line that goes up to the right has a y-intercept of +4. This is where it crosses the y-axis. It's slope (rise/run) is 3/1 = 3, so its equation in slope-intercept form is ...
y = mx +b . . . . where m is the slope, b is the y-intercept
y = 3x +4
The other line has a negative slope and a y-intercept of -1. The slope of that line is rise/run = -2/1 = -2, so its equation is ...
y = -2x -1
__
The solution point will have the x-coordinate that is the solution of the equation ...
y = y
3x +4 = -2x -1 . . . . . . substituting the above expressions for y.
At Jefferson Middle School, eighty-two students were asked which sports they plan to participate in for the coming year. Twenty students plan to participate in track and cross country; six students in cross country and basketball; and eight students in track and basketball. Twelve students plan to participate in all three sports. A total of thirty students plan to participate in basketball, and a total of forty students plan to participate in cross country. Ten students don't plan to participate in any of the three sports. How many students plan to just participate in cross country? 2 4 40 30
Answer:
40
Step-by-step explanation:
In the question only lies the answer:
"and a total of forty students plan to participate in cross country."
Answer:
2
Step-by-step explanation:
2
Complete the table of values for y=-x^2+2x+1
X -3, -2, -1,0,1,2,3,4,5
Y -14,7, ,1, -2 -14
Answer:
see the attachment
Step-by-step explanation:
When you have a number of function evaluations to do, it is convenient to let a graphing calculator or spreadsheet do them. That avoids the tedium and the mistakes in arithmetic.
Here's your completed table.
Simple math! What is the issue with my work? I got it wrong.
Answer:
x = 6
Step-by-step explanation:
In the third line of the solution on right side of the equal sign, middle term should be 8x instead of 4x.
The final value of x will be 6.
[tex] PQ^2 + QO^2 = PO^2 \\
x^2 + 8^2 = (4+x)^2 \\
x^2 + 64 = 16 + 8x + x^2 \\
64 = 16 + 8x \\
64 - 16 = 8x \\
48 = 8x \\
6 = x\\[/tex]
Evaluate 2/3 + 1/3 + 1/6 + …
Answer:
7/6
Step-by-step explanation:
The LCD of these three fractions is 6; the denominators 3, 3 and 6 divide evenly into 6.
Therefore we have:
4/6 + 2/6 + 1/6 = 7/6
PLS HELP :Find all the missing elements:
Answer:
[tex]\large \boxed{\mathrm{34.2}}[/tex]
Step-by-step explanation:
[tex]\sf B= arcsin (\frac{b \times sin(A)}{a} )[/tex]
[tex]\sf B= arcsin (\frac{7 \times sin(40\°)}{8} )[/tex]
[tex]\sf B = 0.59733 \ rad = 34.225\°[/tex]
Find the minimum sample size n needed to estimate for the given values of c, , and E. c, , and E Assume that a preliminary sample has at least 30 members.
Answer:
hello your question is incomplete below is the complete question
Find the minimum sample size n needed to estimate μ For the given values of c, σ, and E. c=0.98, σ=6.5, and E=22 Assume that a preliminary sample has at least 30 members.
Answer : 48
Step-by-step explanation:
Given data:
E = 2.2,
std ( σ ) = 6.5
c ( level of confidence ) = 0.98
To find the minimum sample size
we have to first obtain the value of [tex]Z_{a/2}[/tex]
note : a can be found using this relation :
( 1 - a ) = 0.98 ----- equation 1
a = 1 - 0.98 = 0.02
hence: a/2 = 0.01
This means that P( Z ≤ z ) = 0.99 the value of z can be found using the table of standard normal distribution. from the table the value of z = 2.33
P( Z ≤ 2.33 ) = 0.99
To obtain the sample size n
[tex]n = (\frac{std*z}{E} )^{2}[/tex]
n = [tex](\frac{6.5*2.33}{2.2} )^2[/tex] = (6.88409)^2
Therefore n ≈ 48
1+2x=6x+11 PLS HELP URGENT
Answer:
x = -5/2
Step-by-step explanation:
1+2x=6x+11
Subtract 2x from each side
1+2x-2x=6x-2x+11
1 = 4x+11
Subtract 11 from each side
1-11 = 4x
-10 =4x
Divide by 4
-10/4 = 4x/4
-5/2 =x
Answer:
[tex]\boxed{x=-\frac{5}{2}}\\[/tex]
Step-by-step explanation:
To begin, get the variable on one side of the equation - preferably the left for standard solution notation (for this equation, it is easier to place it on the right side to avoid negative values). Do this by subtracting 2x from both sides of the equation. Then, subtract 11. Finally, divide by 4 and get the answer in terms of x.
1 + 2x = 6x + 11
1 = 4x + 11
-10 = 4x
[tex]\boxed{x=-\frac{5}{2}}[/tex]
According to the Federal Communications Commission, 70% of all U.S. households have vcrs. In a random sample of 15 households, what is the probability that fewer than 13 have vcrs?
Answer:
The probability is [tex]P(x < 13) = 0.8732[/tex]
Step-by-step explanation:
From the question we are told that
The probability of success is p = 0.70
The sample size is [tex]n = 15[/tex]
Generally the distribution of U.S. households have vcrs follow a binomial distribution given that there are only two outcome (household having vcrs or household not having vcrs )
The probability of failure is mathematically evaluated as
[tex]q = 1- p[/tex]
substituting values
[tex]q = 1- 0.70[/tex]
[tex]q = 0.30[/tex]
The probability that fewer than 13 have vcrs is mathematically represented as
[tex]P(x < 13) = 1- [P(13) + P(14) + P(15)][/tex]
=> [tex]P(x < 13) = 1-[( \left 15 } \atop {}} \right. C_{13} *p^{13}* q^{15-13})+ (\left 15 } \atop {}} \right. C_{14} *p^{14}* q^{15-14}) +( \left 15 } \atop {}} \right. C_{15} *p^{15}* q^{15-15}) ][/tex]
Here [tex]\left 15 } \atop {}} \right. C_{13}[/tex] means 15 combination 13 and the value is 105 (obtained from calculator)
Here [tex]\left 15 } \atop {}} \right. C_{14}[/tex] means 15 combination 14 and the value is 15 (obtained from calculator)
Here [tex]\left 15 } \atop {}} \right. C_{15}[/tex] means 15 combination 15 and the value is 1 (obtained from calculator)
So
[tex]P(x < 13) = 1-[(105 *p^{13}* q^{2})+ (15 *p^{14}* q^{1}) +(1*p^{15}* q^{0}) ][/tex]
substituting values
[tex]P(x < 13) = 1-[(105 *(0.70)^{13}* (0.30)^{2})+ (15 *(0.70)^{14}* (0.30)^{1}) +(1*(0.70)^{15}* (0.30)^{0}) ][/tex]
[tex]P(x < 13) = 0.8732[/tex]
A test is being conducted to test the difference between two population means using data that are gathered from a matched pairs experiment. If the paired differences are normal, then the distribution used for testing is the:
Answer:
Student t-distribution.
Step-by-step explanation:
In this scenario, a test is being conducted to test the difference between two population "means" using data that are gathered from a matched pairs experiment. If the paired differences are normal, then the distribution used for testing is the student t-distribution.
In Statistics and probability, a student t-distribution can be defined as the probability distribution which can be used to estimate population parameters when the population variance is not known (unknown) and the sample population is relatively small. The student t-distribution is a statistical distribution which was published in 1908 by William Sealy Gosset.
A student t-distribution has a similar curve with the normal distribution curve, except that it is fatter and a little bit shorter.
