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jek_recluse [69]
2 years ago
12

If 1134 bricks are used to build a wall that is 18ft long, how high is the wall?

Mathematics
1 answer:
Trava [24]2 years ago
6 0
Need some more info, what are the dimensions of the bricks?

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Rearrange x=3g+2 to make g the subject
Eduardwww [97]

Answer:

g = (x - 2)/3

Step-by-step explanation:

g = (x - 2)/3

8 0
2 years ago
Read 2 more answers
A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s
denpristay [2]

Answer:

For this case the  95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

Step-by-step explanation:

Notation

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=40-1=39

For this case the  95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

3 0
2 years ago
What is the standard deviation of the market portfolio if the standard deviation of a fully diversified portfolio with a beta of
Nina [5.8K]

Answer:

22.5%

Step-by-step explanation:

let the standard deviation for market portfolio = σₙ

Also let the standard deviation for fully diversified portfolio = σₓ

<u>To calculate fully diversified portfolio</u>

fully diversified portfolio has <em>σₓ = βσₙ</em>

From the given question beta (β) = 1.25

Also standard deviation for market portfolio (σₙ)  = 18% = 0.18

<em>From the equation above,  σₓ = βσₙ </em>= 1.25×0.18 = 0.225

                                                             = 22.5% (converting to percentage)

<em></em>

3 0
1 year ago
A game is played using one die. If the die is rolled and shows 1, the player wins $5. If the die shows any number other than 1,
mash [69]

Answer:

1/5

Step-by-step explanation:

i had a similar question

For each roll you start with paying 2 dollars and you only with 10 dollars one out of 6 rolls (on average). 

So the cost for one play is 2 dollars and your win is 10/6. 

Value is -2+10/6=-1/3 dollars

So you lose 1/3 dollars on average with each game

since you have no limited rolls u put 1/5

this from another question but both same just different numbers

6 0
2 years ago
The prices of three t-shirts styles are $24, $30 and $36. the probability of choosing a $24 t-shirt is 1/6. the probability of c
Slav-nsk [51]

\text{Answer} : \text{The expected value of a t-shirt is \$31.}

Explanation:

Since we have given that

The prices of three t-shirts styles  i.e $24, $30, $36 with their probability is given by

\frac{1}{6}, \frac{1}{2},\frac{1}{3}

As we know that,

E(X)= \sum_{1}^{3}x_iP(x_i)

\text{where} x_i \text{ is the prices of t- shirts styles}

Now,

x_1= \$24 , x_2=\$30 , x_3=$36

and

P(x_1)=\frac{1}{6},P(x_2)=\frac{1}{2}, P(x_3)=\frac{1}{3}

So,

E(X)= 24\times \frac{1}{6}+30\times\frac{1}{2}+36\times \frac{1}{3}\\=4+15+12\\=31

So, the expected value of a t-shirt = $31.

4 0
2 years ago
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