Joe charges 5 dollar to wash each floor, How many floors does joe need to wash to make at least $32?

ruslelena [56]

ok so lets start of with the fact that the whole thing is 21 units.

So we do 21-9=12 so then we now know that the rectangle is 12 so then we do 12*8 units which is= 96 . Now the triangles. so the first one we know that its 8*3 and times it by 1/2 because two triangles is equal to a rectangle so the first triangle is 12 and now the second. its 9-3= 6 then its 6*8*1/2 which is equal to 24 so now the final answer is

96+12+24 which is equal to 132 so i'm guessing it 132 square units

5 0

3 years ago

2x+3(x-4)=4(2x+3) solving one linear equation form

Art [367]

Answer:

x= -8

Step-by-step explanation:

2x+3x-12=8x+12

5x-12=8x+12

5x-8x=12+12

-3x=24

24/-3

x= -8

4 0

3 years ago

PLEASE HELP VERY URGENT ITS DUE TODAY PLEASE HELP Complete the table for the given rule. Rule: y=1/4x + 1

harkovskaia [24]

Answer: Y = 2 ; 3 ; 4

Step-by-step explanation:

Given the rule : y=1/4x + 1

X = 4, 8, 12

When x = 4

y = 1/4(4) + 1

y = 1 + 1

y = 2

When x = 8

y = 1/4(8) + 1

y = 8/4 + 1

y = 2 + 1

y = 3

When x = 12

y = 1/4(12) + 1

y = 12/4 + 1

y = 3 + 1

y = 4

Therefore ;

X - - - 4 - - - 8 - - - 12

Y - - - 2 - - - 3 - - - 4

5 0

3 years ago

A town’s January high temperatures average 36 degrees (F) with a standard deviation of 10 degrees while in July the mean high

n200080 [17]

Answer:

55 degrees is more unusual in July as it is away from mean temperature as compared to January.

Step-by-step explanation:

We are given the following information in the question:

January:

Mean, μ = 36 degrees

Standard Deviation, σ = 10 degrees

July:

Mean, μ = 74 degrees

Standard Deviation, σ = 8 degrees

We calculate z scores corresponding to x = 55.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

January:

$z_{score} = \displaystyle\frac{55-36}{10} = 1.9$

July:

$z_{score} = \displaystyle\frac{55-74}{8} = -2.375$

Thus, 55 degrees is more unusual in July as it is away from mean temperature as compared to January.

3 0

4 years ago

As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a

REY [17]

Answer:

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

Step-by-step explanation:

Before building the confidence interval we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample of 800 teens. 73% said that they use social networking sites.

This means that:

$p_T = 0.73, s_T = \sqrt{\frac{0.73*0.27}{800}} = 0.0157$

Sample of 2253 adults. 47% said that they use social networking sites.

This means that:

$p_A = 0.47,s_A = \sqrt{\frac{0.47*0.53}{2253}} = 0.0105$

Distribution of the difference:

$p = p_T - p_A = 0.73 - 0.47 = 0.26$

$s = \sqrt{s_T^2+s_A^2} = \sqrt{0.0157^2+0.0105^2} = 0.019$

Confidence interval:

Is given by:

$p \pm zs$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Lower bound:

$p - 1.96s = 0.26 - 1.96*0.019 = 0.223$

Upper bound:

$p + 1.96s = 0.26 + 1.96*0.019 = 0.297$

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

3 0

3 years ago