9514 1404 393
Answer:
(18 1/48)π ≈ 56.61 units
Step-by-step explanation:
Arc length is the product of radius and intercepted arc in radians:
s = rθ
s = (21 5/8)(5π/6) = (18 1/48)π ≈ 56.61 . . . units
Help I don’t get this
Answer:
Step-by-step explanation:
5t² + 4t = 5t² + 4 if and only if t=1.
A zero coefficient makes the value of the term equal to zero.
Answer: 5t to the second power and 5t to the second power. I don't know the answer to b.
Step-by-step explanation:
5 = –6x2 + 24x
5 = –6(x2 – 4x)
inside the parentheses and
.
–19 = –6(x – 2)2
StartFraction 19 Over 6 EndFraction = (x – 2)2
Plus or minus StartRoot StartFraction 19 Over 6 EndFraction EndRoot = x – 2
The two solutions are
Plus or minus StartRoot StartFraction 19 Over 6 EndFraction EndRoot.
Answer:
x = 2 - sqrt(19/6)
x = 2 + sqrt(19/6)
Step-by-step explanation:
Answer:
add 4
subtract 24 from 5
2
Step-by-step explanation:
Find the volume of a cone whose height is 12 and whose radius is 4. Use 3.14 for Pi and round your final answer to one decimal place.
Answer:
201.088 cubic units
Step-by-step explanation:
Given data
height= 12 units
radius= 4 units
The expression for the volume of a cone is given as
Volume = 1/3*πr^2h
Substitute the given data to find the volume of the cone
Volunme= 1/3* 3.142* 4^2* 12
Volume = 1/3* 3.142* 16*12
Volume= 1/3* 603.264
Volume= 201.088 cubic units
What is the area of the figure
Answer:
72 in. sq.
Step-by-step explanation:
6 x 6 = 36
6 x 6 = 36 / 2 = 18
6 x 6 = 36 / 2 = 18
18 + 18 + 36 = 72.
Hope this helps!
If there is something wrong just let me know.
A transect is an archaeological study area that is 1/5 mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance σ2 = 42.3. In a different section of Chaco Canyon, a random sample of 28 transects gave a sample variance s2 = 48.3 for the number of sites per transect.
Required:
Use a 5% level of significance to test the claim that the variance in the new section is greater than 42.3.
Answer:
There is significant evidence to conclude that carina e in New section is greater than 42.3
Step-by-step explanation:
Given :
Sample variance, s² = 48.3
Population variance, σ² = 42.3
Sample size, n = 28
α = 0.05
The hypothesis :
H0 : σ² = 42.3
H0 : σ² > 42.3
The test statistic, χ² : (n-1)*s²/σ²
χ² = [(28 - 1) * 48.3] / 42.3
χ² = (27 * 48.3) / 42.3
χ² = 1304.1 / 42.3
χ² = 30.829787
χ² = 30.830
The Critical value at α = 0.05 ; df = (28-1) = 27
Critical value(0.05, 27) = 27.587
Reject H0 if χ² statistic > Critical value
Since, 30.830 > 27.587 ; Reject H0 ; and conclude that variance in the new section is greater than 42.3
HELP ASAP!!!!!!
Thank you so much
We know
[tex]\boxed{\sf P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]
[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{3(7)+5(-1)}{3+5},\dfrac{3(7)+5(1)}{3+5}\right)[/tex]
[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{21-5}{8},\dfrac{21+5}{8}\right)[/tex]
[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{16}{8},\dfrac{26}{8}\right)[/tex]
[tex]\\ \sf\longmapsto p(x,y)=\left(2,\dfrac{13}{4}\right)[/tex]
m:n=3:5
We know
[tex]\boxed{\sf M(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]
[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{3(7)+5(-1)}{3+5},\dfrac{3(7)+5(1)}{3+5}\right)[/tex]
[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{21-5}{8},\dfrac{21+5}{8}\right)[/tex]
[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{16}{8},\dfrac{26}{8}\right)[/tex]
[tex]\\ \sf\longmapsto M(x,y)=\left(2,\dfrac{13}{4}\right)[/tex]
A man owns 3/4 of the share of a business and sells 1/3 of his
shares for Birr 10,000. What is the value of the business in Bir?
Given:
A man owns 3/4 of the share of a business and sells 1/3 of his shares for Bir 10,000.
To find:
The value of the business in Bir.
Solution:
Let x be the value of the business.
It is given that a man owns 3/4 of the share of a business and sells 1/3 of his shares for Bir 10,000.
[tex]x\times \dfrac{3}{4}\times \dfrac{1}{3}=10000[/tex]
[tex]x\times \dfrac{1}{4}=10000[/tex]
Multiply both sides by 4.
[tex]x=4\times 10000[/tex]
[tex]x=40000[/tex]
Therefore, the value of the business is 40,000 Bir.
please help me with this on the image
Answer:
ik only 1 of them and i dont even know if this is correct...
a) A
A humanities professor assigns letter grades on a test according to the following scheme.
A: Top 8% of scores
B: Scores below the top 8% and above the bottom 62%
C: Scores below the top 38% and above the bottom 18%
D: Scores below the top 82% and above the bottom 9%
E: Bottom 9% of scores Scores on the test are normally distributed with a mean of 67 and a standard deviation of 7.3.
Find the numerical limits for a C grade.
Answer:
The numerical limits for a C grade are 60.6 and 69.1.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Scores on the test are normally distributed with a mean of 67 and a standard deviation of 7.3.
This means that [tex]\mu = 67, \sigma = 7.3[/tex]
Find the numerical limits for a C grade.
Below the 100 - 38 = 62th percentile and above the 18th percentile.
18th percentile:
X when Z has a p-value of 0.18, so X when Z = -0.915.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.915 = \frac{X - 67}{7.3}[/tex]
[tex]X - 67 = -0.915*7[/tex]
[tex]X = 60.6[/tex]
62th percentile:
X when Z has a p-value of 0.62, so X when Z = 0.305.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.305 = \frac{X - 67}{7.3}[/tex]
[tex]X - 67 = 0.305*7[/tex]
[tex]X = 69.1[/tex]
The numerical limits for a C grade are 60.6 and 69.1.
PLEASE HELP ME AND BE CORRECT
Answer:
please someone help me with my latest question
Answer:
41 units
Step-by-step explanation:
Its the reflection the length are same
whats 2 plus 2
*just trying to help someone get points* :)
Answer:4 ma boi
Step-by-step explanation:
Answer:
14
Step-by-step explanation:
because I am god at meth and very smart
A professor, transferred from Toronto to New York, needs to sell his house in Toronto quickly. Someone has offered to buy his house for $220,000, but the offer expires at the end of the week. The professor does not currently have a better offer but can afford to leave the house on the market for another month. From conversations with his realtor, the professor believes the price he will get by leaving the house on the market for another month is uniformly distributed between $210,000 and $235,000. If he leaves the house on the market for another month, what is the probability that he will get at least $225,000 for the house
the line that passes through the point (-4, 2) and has a
What is the equation of
slope of
2?
Answer:
y = 2x + 10
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Here m = 2 , then
y = 2x + c ← is the partial equation
To find c substitute (- 4, 2 ) into the partial equation
2 = - 8 + c ⇒ c = 2 + 8 = 10
y = 2x + 10 ← equation of line
Open the graphing tool one last time. Compare the graphs of y=log (x-k) and y=log x+k in relation to their domain, range, and asymptotes. Describe what you see.
Answer:
sorry I don't know the answer
Answer:
For the equation y=log(x-k), the domain depends on the value of K. Sliding K moves the left bound of the domain interval. The range and the right end behavior stay the same. For the equation y=log x+k, the domain is fixed, starting at an x-value of 0. The vertical asymptote is also fixed. The range of the equation depends on K.
Step-by-step explanation:
Select the correct answer. Which graph represents this inequality? y ≥ 4x − 3
Step-by-step explanation:
You didn't put the graph, but you can compare between your graphs and the picture.
Brainliest please
The graph that represents this inequality y ≥ 4x − 3 is attached below.
What is a solution set to an inequality or an equation?If the equation or inequality contains variable terms, then there might be some values of those variables for which that equation or inequality might be true. Such values are called solution to that equation or inequality. Set of such values is called solution set to the considered equation or inequality.
We are given that the inequality is;
y ≥ 4x − 3
The slope of the inequality is 4.
The equation of the red line is y = 4x − 3
The shading is above the line and the line is solid, that means y is greater than or equal 4x − 3
The graph of this inequality y ≥ 4x − 3 is attached below.
Learn more about inequalities here:
https://brainly.com/question/27425770
#SPJ2
distance between 4, -4 and -7, -4
Step-by-step explanation:
here's the answer to your question
Answer: Distance = 11
Step-by-step explanation:
Concept:
Here, we need to know the idea of the distance formula.
The distance formula is the formula, which is used to find the distance between any two points.
If you are still confused, please refer to the attachment below for a clear version of the formula.
Solve:
Find the distance between A and B, where:
A (4, -4)B (-7, -4)[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]Distance=\sqrt{(4+7)^2+(-4+4)^2}[/tex]
[tex]Distance=\sqrt{(11)^2+(0)^2}[/tex]
[tex]Distance=\sqrt{121+0}[/tex]
[tex]Distance=\sqrt{121}[/tex]
[tex]Distance=11[/tex]
Hope this helps!! :)
Please let me know if you have any questions
https://www.drfrostmaths.com/util-generatekeyskillpic.php?name=AnglePoint4&width=400¶ms=%5B61%2C0%2C%22y%22%2C4%5D
Answer:
Complementary angle
<y+61° =90°
<y = 90°- 61°
<y =29°
I Hope this helps.
a gym class has 10 boys and 12 girls. how many ways can a team of 6 be selected if the team must have the same number of boys and girls
Answer:
The number of ways of selecting the team is 26,400 ways.
Step-by-step explanation:
Given;
total number boys in the gym, b = 10 boys
total number of girls in the gym, g = 12 girls
number of team to be selected, n = 6
If there must equal number of boys and girls in the team, then the team must consist of 3 boys and 3 girls.
Number of ways of choosing 3 boys from the total of 10 = [tex]10_C_3[/tex]
Number of ways of choosing 3 girls from a total of 12 = [tex]12_C_3[/tex]
The number of ways of combining the two possibilities;
[tex]n = 10_C_3 \times 12_C_3\\\\n = \frac{10!}{7!3!} \ \times \ \frac{12!}{9!3!} \\\\n = \frac{10\times 9 \times 8}{3\times 2} \ \times \ \frac{12\times 11 \times 10}{3\times 2} \\\\n = 120 \times 220\\\\n = 26,400 \ ways[/tex]
Therefore, the number of ways of selecting the team is 26,400 ways.
URGENT HELP
The gradient of the tangent to the curve y = ax + bx^3 at the point (2, -4) is 6.
Determine the unknowns a and b.
a=?
b=?
Answer:
a = -6
b = 1
Step-by-step explanation:
The gradient of the tangent to the curve y = ax + bx^3, will be:
dy/dx = a + 3bx²
at (2, -4)
dy/dx = a+3b(2)²
dy/dx = a+12b
Since the gradient at the point is 6, then;
a+12b = 6 ....1
Substitute x = 2 and y = -4 into the original expression
-4 = 2a + 8b
a + 4b = -2 ...2
a+12b = 6 ....1
Subtract
4b - 12b = -2-6
-8b = -8
b = -8/-8
b = 1
Substitute b = 1 into equation 1
Recall from 1 that a+12b = 6
a+12(1) = 6
a = 6 - 12
a = -6
Hence a = -6, b = 1
Student scores on exams given by a certain instructor have mean 74 and standard deviation 14. This instructor is about to give two exams, one to a class of size 25 and the other to a class of size 64.
Required:
a. Approximate the probability that the average test score in the class of size 25 exceeds 80.
b. Repeat part (a) for the class of size 64.
c. Approximate the probability that the average test score in the larger class exceeds that of the other class by over 2.2 points.
Answer:
a) Hence the probability that the average test score in the class of size 25 exceeds 80.
P ( X > 74) = 0.9838
P ( Z > 2.14) = 0.0162
b) Hence the probability that the average test score for the class of size 64
P ( X > 74) = 0.9838
P ( Z > 2.14) = 0.0003
c) Probability of the difference exceeding 2.2 = 0.9936
P (Z < 2.49) = 0.0064
Step-by-step explanation:
Let's assume a normal distribution.
Now,
a) For a class of 25
[tex]P ( X > 74) = P (Z > 80 - 74/ 14\sqrt(25) )\\\\= P (Z < 6 / 2.8)\\\\= P ( Z < 2.14)\\= 0.9838\\[/tex]
[tex]P ( Z > 2.14) = 1- 0.9838\\= 0.0162[/tex]
b)
Similarly:
For the class of 64
[tex]P ( X > 74) = P (Z > 80 - 74/ 14\sqrt(64) )\\\\= P (Z < 6 / 1.75)= P ( Z < 3.428)\\= 0.9838\\[/tex]
[tex]P ( Z > 2.14) = 1- 0.9997\\= 0.0003[/tex]
c) Probability of the difference exceeding 2.2
[tex]= P (Z > 2.2/\sqrt{14 * {(1/25) + (1/64}\\[/tex]
[tex]= P (Z > 2.49)\\= 0.9936[/tex]
P (Z < 2.49)
= 1 - 0.9936
= 0.0064
А _______ equation can be written in the form ax2 + bx+c=0 where a, b, and c are real numbers, and a is a nonzero number.
Fill in the blank.
A) quadratic
B) quartic
C) linear
D) cubic
Wrong answers WILL be reported. Thanks!
Answer:
A) quadratic
Step-by-step explanation:
ax2 + bx+c=0
Since the highest power of the equation is 2
A) quadratic -2
B) quartic- 4
C) linear- 1
D) cubic-3
Use the procedures developed to find the general solution of the differential equation. (Let x be the independent variable.)
2y''' + 15y'' + 24y' + 11y= 0
Solution :
Given :
2y''' + 15y'' + 24y' + 11y= 0
Let x = independent variable
[tex](a_0D^n + a_1D^{n-1}+a_2D^{n-2} + ....+ a_n) y) = Q(x)[/tex] is a differential equation.
If [tex]Q(x) \neq 0[/tex]
It is non homogeneous then,
The general solution = complementary solution + particular integral
If Q(x) = 0
It is called the homogeneous then the general solution = complementary solution.
2y''' + 15y'' + 24y' + 11y= 0
[tex]$(2D^3+15D^2+24D+11)y=0$[/tex]
Auxiliary equation,
[tex]$2m^3+15m^2+24m +11 = 0$[/tex]
-1 | 2 15 24 11
| 0 -2 - 13 -11
2 13 11 0
∴ [tex]2m^2+13m+11=0[/tex]
The roots are
[tex]$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$[/tex]
[tex]$=\frac{-13\pm \sqrt{13^2-4(11)(2)}}{2(2)}$[/tex]
[tex]$=\frac{-13\pm9}{4}$[/tex]
[tex]$=-5.5, -1$[/tex]
So, [tex]m_1, m_2, m_3 = -1, -1, -5.5[/tex]
Then the general solution is :
[tex]$= (c_1+c_2 x)e^{-x} + c_3 \ e^{-5.5x}$[/tex]
Which graph shows the solution to the given system of inequalities? [y<6x+1 y<-3.2x-4
Answer:
VERY NICE RACK U HAVE MAM
Step-by-step explanation:
Answer:
Its a
Step-by-step explanation:
found on another thing and im taking test
3w2 – 21w = 0
Need some help.
Answer:
The solutions are w=0 ,7
Step-by-step explanation:
3w^2 – 21w = 0
Factor out 3w
3w(w-7) =0
Using the zero product property
3w=0 w-7=0
w =0 w=7
The solutions are w=0 ,7
Question 7 of 13
Find the solution to the system of equations,
5x - 3y - Z= 6
-4x + 5y + z = 6
Х
+ 3z = 10
Answer:
D. x = 4, y = 4, z = 2
Step-by-step explanation
Plugged in given answers as trying to substitute is impossible, already tried all combinations
The sum of two binomials is 12x2 − 5x. If one of the binomials is x2 − 2x, the other binomial is:
1. 11x2 − 7x.
2. 12x2 − 3x.
3. 11x2 − 3x.
4. None of these choices are correct.
Answer:
C. 11x² - 3x
Step-by-step explanation:
(12x² - 5x) - (x² - 2x)
12x² - 5x - x² + 2x
12x - x² - 5x + 2x
11x² - 3x
Can someone help me with this problem
Answer:
3/11
Step-by-step explanation:
Solve 5 to x-4 power =7
Answer:The value of x is 4.
Step-by-step explanation:
Given : Expression .
To find : Solve for x ?
Solution :
Applying exponential property,
Comparing the base,
Step-by-step explanation:
What is the numerical coefficient of the first term
Answer:
the number before the first variable (first term)
Step-by-step explanation:
this appears to be an incomplete question. The numerical coefficient of a term is the number before the variable.
the constant is the number without a variable.
What is a1
of the arithmetic sequence for which a3=126
and a64=3,725
a
64
=
3
,
725
?
In an arithmetic sequence, every pair of consecutive terms differs by a fixed number c, so that the n-th term [tex]a_n[/tex] is given recursively by
[tex]a_n=a_{n-1}+c[/tex]
Then for n ≥ 2, we have
[tex]a_2=a_1+c[/tex]
[tex]a_3=a_2+c = (a_1+c)+c = a_1 + 2c[/tex]
[tex]a_4=a_3+c = (a_1 + 2c) + c = a_1 + 3c[/tex]
and so on, up to
[tex]a_n=a_1+(n-1)c[/tex]
Given that [tex]a_3=126[/tex] and [tex]a_{64}=3725[/tex], we can solve for [tex]a_1[/tex]:
[tex]\begin{cases}a_1+2c=126\\a_1+63c=3725\end{cases}[/tex]
[tex]\implies(a_1+63c)-(a_1+2c)=3725-126[/tex]
[tex]\implies 61c = 3599[/tex]
[tex]\implies c=59[/tex]
[tex]\implies a_1+2\times59=126[/tex]
[tex]\implies a_1+118 = 126[/tex]
[tex]\implies \boxed{a_1=8}[/tex]
