Focus on Gamble B only. Multiply each winnings with their corresponding probabilities.
5100*0.50 = 2550
200*0.25 = 50
0*0.25 = 0
Add up those results: 2550+50+0 = 2600
The expected value of gamble B is $2600
What is the slope of the line with the two
points A(-4, 8) and B(-9, 12)?
Answer:
slope = -4/5
Step-by-step explanation:
A line passes two points (x1, y1) and (x2, y2).
The slope of this line can be calculate by the formula:
s = (y2 - y1)/(x2 - x1)
=>The line that passes A(-4, 8) and B(-9, 12) has the slope:
s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5
Hope this helps!
I need help for the solution
Answer:
[tex]\boxed{ \ dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t\ }[/tex]
Step-by-step explanation:
it is a long time I have not applied Ito's lemma
I would say the following
for [tex]f(x)=x^2[/tex]
f'(x)=2x
f''(x)=2
so using Ito's lemma we can write that
[tex]dY_t=2V_tdV_t+\phi^2dt[/tex]
[tex]dY_t=2(\theta+\psi V_t^2)dt+2\phi V_tdW_t+\phi^2dt[/tex]
[tex]dY_t=(2\theta+2\psi V_t^2+\phi^2)dt+2\phi V_tdW_t[/tex]
so it comes
[tex]dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t[/tex]
Find the area of the smaller sector.
A
6 in
030°
Area = [? ]in?
B
Round your answer to the nearest hundredth.
Answer:
9.42 in²
Step-by-step explanation:
The area of whole circle S=pi*R² , where pi is appr. 3.14, R= 6 in
S= 3.14*6² =113.04 in²
The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²
The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.
To find the area of a sector, you can use the formula:
Area of sector = (θ/360) × π × r²
where θ is the central angle in degrees, r is the radius of the sector.
The central angle is 30 degrees and the radius is 6 inches.
Plugging these values into the formula:
Area of sector = (30/360) × π × 6²
= (1/12) × π × 36
= (1/12) × 3.14159 × 36
= 9.42478 square inches
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Consider random samples of size 900 from a population with proportion 0.75 . Find the standard error of the distribution of sample proportions. Round your answer for the standard error to three decimal places. standard error
Answer:
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] n=900[/tex] represent the sample size selected
[tex]p = 0.75[/tex] represent the population proportion
We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:
[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]
And the standard error is given;
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]
ali's typing rate between 8:00 am and noon is 48 words per minute . after lunch a lunch break, Ali's typing rate between 1:00 pm and 4:00 pm is 2,040 words per hour . what is Ali's average typing rate per minute for the whole time she works?
Answer:
41 word/min
Step-by-step explanation:
Before noon Ali works:
4 hours= 4*60 min= 240 minShe types:
240*48= 11520 wordsAfter lunch she works:
4 hoursShe types:
4*2040= 8160 wordsTotal Ali works= 4+4= 8 hours= 480 min
Total Ali types= 11520+8160= 19680 words
Average typing rate= 19680 words/480 min= 41 word/min
The production department has installed a new spray machine to paint automobile doors. As is common with most spray guns, unsightly blemishes often appear because of improper mixtures or other problems. A worker counted the number of blemishes on each door. Most doors had no blemishes; a few had one; a very few had two; and so on. The average number was 0.5 per door. The distribution of blemishes followed the Poisson distribution. Out of 10,000 doors painted, about how many would have no blemishes
Answer:
The numbers of doors that will have no blemishes will be about 6065 doors
Step-by-step explanation:
Let the number of counts by the worker of each blemishes on the door be (X)
The distribution of blemishes followed the Poisson distribution with parameter [tex]\lambda =0.5[/tex] / door
The probability mass function on of a poisson distribution Is:
[tex]P(X=x) = \dfrac{e^{- \lambda } \lambda ^x}{x!}[/tex]
[tex]P(X=x) = \dfrac{e^{- \ 0.5 }( 0.5)^ x}{x!}[/tex]
The probability that no blemishes occur is :
[tex]P(X=0) = \dfrac{e^{- \ 0.5 }( 0.5)^ 0}{0!}[/tex]
[tex]P(X=0) = 0.60653[/tex]
P(X=0) = 0.6065
Assume the number of paints on the door by q = 10000
Hence; the number of doors that have no blemishes is = qp
=10,000(0.6065)
= 6065
pls helppppp with my math
Answer:
[tex]\frac{1}{6}[/tex]
Step-by-step explanation:
Answer:
Step-by-step explanation:
[tex]\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{2*2}{3*2}\\\\=\frac{5}{6}-\frac{4}{6}\\\\=\frac{5-4}{6}\\\\=\frac{1}{6}[/tex]
The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a weekend is $600 or less. A member of the hotel's accounting staff noticed that the total charges for guest bills have been increasing in recent months. The accountant will use a sample of future weekend guest bills to test the manager's claim. (a) Which form of the hypotheses should be used to test the manager's claim? H0: - Select your answer - Ha: - Select your answer - The member of the hotel's accounting staff suspects that the total charges for guest bills have Select in recent months. To test the manager’s claim, the staff member will conduct Select test of the population Select . (b) What conclusion is appropriate when H0 cannot be rejected? When H0 cannot be rejected, there Select enough evidence to conclude that the total charges for guest bills have Select in recent months. (c) What conclusion is appropriate when H0 can be rejected? When H0 can be rejected, there Select enough evidence to conclude that the total charges for guest bills have Select in recent m
Answer:
a) Null hypothesis (H0): the mean guest bill for a weekend is $600.
Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.
b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.
c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.
Step-by-step explanation:
a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:
Null hypothesis (H0): the mean guest bill for a weekend is $600.
Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.
A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.
b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.
c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.
Pet Place sells pet food and supplies including a popular bailed hay for horses. When the stock of this hay drops to 20 bails, a replenishment order is placed. The store manager is concerned that sales are being lost due to stock outs while waiting for a replenishment order. It has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails. The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 bails
Answer:
The probability that demand during lead-time will exceed 20 bails is 0.2033.
Step-by-step explanation:
We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.
Let X = demand during the lead-time
So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu=[/tex] population mean demand = 15 bails
[tex]\sigma[/tex] = standard deviation = 6 bails
Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)
P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)
= 1 - 0.7967 = 0.2033
Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n
Answer:
Option C is correct.
The sampling distribution with sample size n=100 will have less variability.
Step-by-step explanation:
Complete Question
Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n = 100 or a sample size of n = 20.
A. Both sampling distributions will have the same variability.
B.The sampling distribution with sample size n=20 will have less variability
C. The sampling distribution with sample size n =100 will have less variability
Solution
The central limit theorem allows us to say that as long as
- the sample is randomly selected from the population distribution with each variable independent of each other and with the sample having an adequate enough sample size.
- the random sample is normal or almost normal which is guaranteed if the population distribution that the random sample was extracted from is normal or approximately normal,
1) The mean of sampling distribution (μₓ) is approximately equal to the population mean (μ)
μₓ = μ = 66 inches
2) The standard deviation of the sampling distribution or the standard error of the sample mean is related to the population standard deviation through
σₓ = (σ/√N)
where σ = population standard deviation = 3.5 inches
N = Sample size
And the measure of variability for a sampling distribution is the magnitude of the standard deviation of the sampling distribution.
For sampling distribution with sample size n = 100
σₓ = (3.5/√100) = 0.35 inch
For sampling distribution with sample size n = 20
σₓ = (3.5/√20) = 0.7826 inch
The standard deviation of the sampling distribution with sample size n = 20 is more than double that of the sampling distribution with sample size n = 100, hence, it is evident that the bigger the sample size, the lesser the standard deviation of the sampling distribution and the lesser the variability that the sampling distribution shows.
Hope this Helps!!!
Please answer this question I give brainliest thank you! Number 8
Answer:
The third options
Step-by-step explanation:
Counting we can see that 10 students went to two or less states, and 10 went to three or more
A 12 ft ladder leans against the side of a house. The top of the ladder is 10ft off the ground. Find x, the angle of elevation of the ladder.
1. Remember to address each of the critical elements of the prompt:
Articulate your overall approach to solving this problem before tackling the details. In other words, think about what the question is actually asking, which pieces of information are relevant, and how you can use what you have learned to fill in the missing pieces.
2. Apply the mathematical process to solve the problem:
Interpret the word problem to identify any missing information.
Translate the word problem into an equation.
Appropriately use the order of operations and law of sines and cosines to determine the solution.
Check your work by ensuring that the known properties of triangles are met.
The image is missing, so i have attached it.
Answer:
x = 56.44°
Step-by-step explanation:
From the attached image, we can see that this is a right angle triangle which has opposite, adjacent and hypotenuse as sides. Since we want to find the angle x, thus, we can make use of trigonometric ratios.
From the attached image, the side opposite to angle x is 10ft and the hypotenuse is 12 ft.
From trigonometric ratios, we know that, sin x = opposite/hypotenuse
So, sin x = 10/12
x = sin^(-1) (10/12)
x = sin^(-1) 0.8333
x = 56.44°
Find the Laplace transform F(s)=L{f(t)} of the function f(t)=sin2(wt), defined on the interval t≥0. F(s)=L{sin2(wt)}= help (formulas) Hint: Use a double-angle trigonometric identity. For what values of s does the Laplace transform exist? help (inequalities)
The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .
The Laplace transform exist when s > 0 .
Here, the given function is f(t) = sin²(wt) .
The Laplace transform of the the function f(t),
F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }
F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }
F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }
F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]
Next,
The above Laplace transform exist if s > 0 .
Know more about Laplace transform,
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A laptop computer is purchased for $2300. Each year, its value is 75% of its value the year before. After how many years will the laptop computer be worth $700 or less? (Use the calculator provided if necessary.) Write the smallest possible whole number answer.
Answer:
after the 1st year
Step-by-step explanation:
$2300 × 75% = $1725.00
$2300-$1725= $575
Suppose that four microchips in a production run of sixty are defective. A sample of six is to be selected to be checked for defects. (a) How many different samples can be chosen
Answer:
50,063,860 different samples can be chosen
Step-by-step explanation:
The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
How many different samples can be chosen
We choose 6 microchips from a set of 60. So
[tex]C_{60,6} = \frac{60!}{6!(60-6)!} = 50063860[/tex]
50,063,860 different samples can be chosen
The mean number of hours of part-time work per week for a sample of 317 teenagers is 29. If the margin of error for the population mean with a 95% confidence interval is 2.1, construct a 95% confidence interval for the mean number of hours of part-time work per week for all teenagers.
Answer:
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex]
[tex]29+2.1=31.1[/tex]
The 95% confidence interval would be between 26.9 and 31.1
Step-by-step explanation:
Information given
[tex]\bar X= 29[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s represent the sample standard deviation
[tex] ME= 2.1[/tex] represent the margin of error
n represent the sample size
Solution
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
And this formula is equivalent to:
[tex] \bar X \pm ME[/te]x
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex]
[tex]29+2.1=31.1[/tex]
The 95% confidence interval would be between 26.9 and 31.1
Two thousand dollars is deposited into a savings account at 8.5% interest compounded continuously. (a) What is the formula for A(t), the balance after t years? (b) What differential equation is satisfied by A(t), the balance after t years? (c) How much money will be in the account after 5 years? (d) When will the balance reach $3000? (e) How fast is the balance growing when it reaches $3000?
Answer:
a)[tex]A(t)=2000e^{0.085t}[/tex]
b)[tex]A'(t)=170e^{0.085t}[/tex]
c)$3059.1808
d)t=4.77 years
e) The balance growing is $254.99/year
Step-by-step explanation:
We are given that Two thousand dollars is deposited into a savings account at 8.5% interest compounded continuously.
Principal = $2000
Rate of interest = 8.5%
a) What is the formula for A(t), the balance after t years?
Formula [tex]A(t)=Pe^{rt}[/tex]
So,[tex]A(t)=2000e^{0.085t}[/tex]
B)What differential equation is satisfied by A(t), the balance after t years?
So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]
[tex]A'(t)=170e^{0.085t}[/tex]
c)How much money will be in the account after 5 years?
Substitute t = 5 in the formula "
[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]
d)When will the balance reach $3000?
Substitute A(t)=3000
So, [tex]3000=2000e^{0.085t}[/tex]
t=4.77
The balance reach $3000 in 4.77 years
e)How fast is the balance growing when it reaches $3000?
Substitute the value of t = 4.77 in derivative formula :
[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]
Hence the balance growing is $254.99/year
The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.5 ppm and standard deviation 1.4 ppm. 18 randomly selected large cities are studied. Round all answers to two decimal places.
A. xBar~ N( ____) (____)
B. For the 18 cities, find the probability that the average amount of pollutants is more than 9 ppm.
C. What is the probability that one randomly selected city's waterway will have more than 9 ppm pollutants?
D. Find the IQR for the average of 18 cities.Q1 =
Q3 =
IQR:
2. X ~ N(30,10). Suppose that you form random samples with sample size 4 from this distribution. Let xBar be the random variable of averages. Let ΣX be the random variable of sums. Round all answers to two decimal places.
A. xBar~ N(___) (____)
B. P(xBar<30) =
C. Find the 95th percentile for the xBar distribution.
D. P(xBar > 36)=
E. Q3 for the xBar distribution =
Answer:
1)
A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552
C) P(X> 9)= 0.36317
D) IQR= 0.4422
2)
A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)
B) P( [tex]\frac{}{X}[/tex]<30)= 0.50
C) P₉₅= 32.60
D) P( [tex]\frac{}{X}[/tex]>36)= 0
E) Q₃: 31.0586
Step-by-step explanation:
Hello!
1)
The variable of interest is
X: pollutants found in waterways near a large city. (ppm)
This variable has a normal distribution:
X~N(μ;σ²)
μ= 8.5 ppm
σ= 1.4 ppm
A sample of 18 large cities were studied.
A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
The population mean is the same as the mean of the variable
μ= 8.5 ppm
The standard deviation is
σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108
So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B)
P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)
To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.
Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]
Then using the Z table you'll find the probability of
P(Z≤1.51)= 0.93448
Then
1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552
C)
In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:
P(X> 9)= 1 - P(X ≤ 9)
Z= (X-μ)/δ= (9-8.5)/1.44
Z= 0.347= 0.35
P(Z≤0.35)= 0.63683
Then
P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317
D)
The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:
Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25
Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:
P(Z≤z₁)= 0.25
Using the table you have to identify the value of Z that accumulates 0.25 of probability:
z₁= -0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)
z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)
[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5= 8.2789 ppm
The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
Using the table you have to identify the value of Z that accumulates 0.75 of probability:
z₃= 0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5= 8.7211 ppm
IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422
2)
A)
X ~ N(30,10)
For n=4
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
Population mean μ= 30
Population variance σ²/n= 10/4= 2.5
Population standard deviation σ/√n= √2.5= 1.58
[tex]\frac{}{X}[/tex] ~ N(30;2.5)
B)
P( [tex]\frac{}{X}[/tex]<30)
First you have to standardize the value and then look for the probability:
Z= ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0
P(Z<0)= 0.50
Then
P( [tex]\frac{}{X}[/tex]<30)= 0.50
Which is no surprise since 30 y the value of the mean of the distribution.
C)
P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95
P( Z≤ z₀)= 0.95
z₀= 1.645
Now you have to reverse the standardization:
z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)
z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)
[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60
P₉₅= 32.60
D)
P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0
E)
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
z₃= 0.67
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586
Q₃: 31.0586
Number of multiples of 7 between 200 and 1000
Answer:
114
Step-by-step explanation:
Answer:
144Step-by-step explanation:
6. Factor the expression.
9b2 + 48b + 64
A (3b + 8)2
B (-3b + 8)2
C (-3b - 82
D (3b - 8)2
70%
Answer:
A. [tex](3b+8)^2[/tex]
Step-by-step explanation:
[tex]9b^2+48b +64\\=(3b)^2 + 2\times 3b\times 8 +(8)^2\\=(3b+8)^2[/tex]
What is the result of −18⋅16 2/3? Enter the result as an improper fraction and as a mixed number.
Answer:
-30000/100
300 0/1
Step-by-step explanation:
We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.
16 2/3 = 16.67
We do the multiplication:
−18⋅16 2/3 = -300
We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:
-30000/100
How mixed number would it be:
300 0/1
A research organization keeps track of what citizens think is the most important problem facing the country today. They randomly sampled a number of people in 2003 and again in 2009 using a different random sample of people in 2009 than in 2003 and asked them to choose the most important problem facing the country today from the following choices, war, economy, health care, or other. Which of the following is the correct test to use to determine if the distribution of "problem facing this countrytoday" is different between the two different years?
A.
Use a chi-square test of homogeneity.
B.
Use a paired t-test.
C.
Use a two-sample z-test for proportions.
D.
Use a chi-square goodness-of-fit test.
Answer:
Step-by-step explanation:
From the information given, the population is divided into sub groups. Each group would consist of citizens picking a particular choice as the most important problem facing the country. The choices are the different categories. In this case, the null hypothesis would state that the distribution of proportions for all categories is the same in each population. The alternative hypothesis would state that the distributions is different. Therefore, the correct test to use to determine if the distribution of "problem facing this country today" is different between the two different years is
A) Use a chi-square test of homogeneity.
Find the following measure for this figure.
Volume =
Answer:
91 2/3 pi cubic units
Step-by-step explanation:
The formula for the volume of cone is [tex]\dfrac{1}{3}\pi r^2 h[/tex]. Plugging in the given numbers, you get:
[tex]\dfrac{1}{3}\cdot \pi \cdot 5^2 \cdot 11= 91 \ 2/3 \pi[/tex]
Hope this helps!
Answer:
[tex]Volume=\frac{1}{3} \,275\,\pi[/tex] cubic units
Notice that this answer doesn't agree with any of the first three in the list provided via the screenshot
Step-by-step explanation:
Recall the formula for the volume of a cone:
[tex]Volume=\frac{1}{3} Base\,*\,Height[/tex]
In this case the Height is 11 units, and they also give us the radius of the circular base (5 units) from which we can find the circle's base area:
[tex]Area_{circle} = \pi\,R^2\\Area_{circle}=\pi\,(5)^2\\Area_{circle}=25 \pi[/tex]
therefore the total volume becomes:
[tex]Volume=\frac{1}{3} Base\,*\,Height\\Volume=\frac{1}{3} 25\,\pi\,*\,11\\\\Volume=\frac{1}{3} \,275\,\pi[/tex]
PLEASE ANSWER FAST !!!
What is the range of the function g for given domain ?
Answer:
The answer is B
Step-by-step explanation:
Hope this helps.. if not im sorry :(
Someone flips five coins, but you don’t see the outcome. The person reports that no tails are showing. What is the probability that the person flipped 5 heads?
Answer:
0.03125 = 3.125% probability that the person flipped 5 heads
Step-by-step explanation:
For each coin, there are only two possible outcomes. Either it was heads, or it was tails. The result of a coin toss is independent of other coin tosses. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Five coins:
This means that n = 5.
Fair coin:
Equally as likely to be heads or tails, so p = 0.5.
What is the probability that the person flipped 5 heads?
This is P(X = 5).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125[/tex]
0.03125 = 3.125% probability that the person flipped 5 heads
The World Issues club has decided to donate 60% of all their fundraising activities this year to Stephen Lewis Foundation. This foundation was created to help ease the pain of HIV/AIDS in Africa. Lewis, a Canadian, works for the United Nations trying to determine ways to stop the spread of this deadly disease from crippling an entire continent. Choose a variable to represent the money earned during fundraising activities and the revenue generated for the foundation Use these variables to create an equation that will determine the amount of money the foundation will receive. In their latest bake sale, the club raised $72. Calculate the amount the foundation will receive. At the end of the year, the World Issues Club mailed a cheque to the foundation for $850. How much money did they fundraise in total?
Answer:
$43.20$1416.67Step-by-step explanation:
Let the money earned during fundraising activities =x
Since the World Issues club has decided to donate 60% of all their fundraising activities this year to Stephen Lewis Foundation.
The amount of money the foundation will receive
=60% of x
= 0.6x
In the bake sale, the club raised $72.
Therefore, the amount the foundation will receive =0.6*72=$43.20
At the end of the year, the World Issues Club mailed a cheque to the foundation for $850.
Therefore:
0.6x=850
x=850/0.6
x=$1416.67
The total amount of money the club raised is $1416.67.
Find the area of a triangle that has the base of 5 inches and a height of 3 3/4 inches
Answer:
9.375 in^2
Step-by-step explanation:
The equation of a circle is x2 + y2 = 56.25. Find the radius of the circle?
Answer:
r = 7.5
Step-by-step explanation:
Circle equation: [tex](x - h)^2 + (y - k)^2 = r^2[/tex]
Since we are already give r², we simply just take the square root of 56.25, and we should get 7.5 as our final answer!
An Undergraduate Study Committee of 6 members at a major university is to be formed from a pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is the probability that precisely half of the members will be women?
Answer:
5/33649= approx 0.00015
Step-by-step explanation:
Total number of outcomes are C24 6= 24!/(24-6)!/6!=19*20*21*22*23*24/(2*3*4*5*6)= 19*14*22*23
Half of the Committee =3 persons. That mens that number of the women in Commettee=3. 3 women from 6 can be elected C6 3 ways ( outputs)=
6!/3!/3!=4*5*6*/2/3=20
So the probability that 3 members of the commettee are women is
P(women=3)= 20/(19*14*22*23)=5/(77*19*23)=5/33649=approx 0.00015
The probability that precisely half of the members will be women is;
P(3 women) = 0.1213
This question will be solved by hypergeometric distribution which has the formula;
P(x) = [S_C_s × (N - S)_C_(n - s)]/(NC_n)
where;
S is success from population
s is success from sample
N is population size
n is sample size
We are give;
s = 3 women (which is precisely half of the members selected)
S = 6 women
N = 24 men and women
n = 6 people selected
Thus;
P(3 women) = (⁶C₃ * ⁽¹⁸⁾C₍₃₎)/(²⁴C₆)
P(3 women) = (20 * 816)/134596
P(3 women) = 0.1213
Read more at; https://brainly.com/question/5733654
Black walnut trees contain chemicals that inhibit the growth of other plants. In a simple experiment to test whether this is true, you grow several tomato plants in soil with and without decomposing leaves from a black walnut tree. You collect data on plant height as a measure of growth. In this experiment, __________ is the independent variable, __________ is the dependent variable, and __________ is the control.
Answer:
Height of tomato plant is the dependent variable
Presence of walnut leaves in the soil is the independent variable
Tomato plants grown without walnut leaves is the control
Step-by-step explanation:
An independent variable is the variable in an experiment that can be altered to test for a certain result. It is independent, or does not change with change in other factors in the experiment. In this case, the presence or absence, or quantity of walnut available in the soil is the independent variable in the experiment.
A dependent variable varies, and depends on the independent variable. It is what is measured in the experiment. In this case, the height of the tomato plants is the dependent variable that depends on the presence, absence or quantity of walnut in the soil.
A control in an experiment, is a replicate experiment, that is manipulated in order to be able to test a single variable at a time. Controls are variables are held constant so as to minimize their effect on the system under study. In this case, some of the tomato plants are planted without walnut in the soil, to test the effect of the absence of the walnut in the soil.