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3 years ago

If you place a 89-foot ladder against the top of a 80-foot building, how many feet will

Mathematics

1 answer:

Nuetrik [128]3 years ago

8 0

Answer:

$\large \boxed{\text{39 ft}}$

Step-by-step explanation:

Let x be the distance from the foot of the ladder to the base of the building.

We have a right triangle, so we can use Pythagoras' Theorem.

$\begin{array}{rcl}80^{2} + x^{2}& = & 89^{2} \\6400+ x^{2} & = & 7921\\x^{2} & = & 1521\\x & = & \sqrt{1521}\\& = &\mathbf{39}\\\end{array}\\\text{The foot of the ladder will be $\large \boxed{\textbf{39 ft}}$ from the base of the building,}$

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Solve for n, 4 x n = 7 x 16

Sever21 [200]

Answer:

n=28 7 times 16=112divided by 4 =28

Step-by-step explanation:

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3 years ago

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Marcia has a licorice rope that is 28 inches long. She wants to cut it into 3,-inch pieces to share with friends. How many piece

diamong [38]

Answer:

9 pieces

Step-by-step explanation:

28 inches rope cut into 3-inch pieces

28/3 = 9 1/3

There is no matching answer choice, if given numbers are correct then the answer is 9 full pieces

---------

The close answer choice is 8
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So if cut into 3.5 inch pieces then it is 8 full pieces
just in case it was typo 3 instead of 3.5

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4 years ago

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Answer:

15/8

Step-by-step explanation:

because it will have to be it because

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3 years ago

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Write each sentence as an equation or inequality. Use x to represent

mash [69]

Answer:

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Step-by-step explanation:

unknown number = x

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3 years ago

A fitness center is interested in finding a 90% confidence interval for the mean number of days per week that Americans who are

Blababa [14]

Answer:

A. Normal

B. Between 2.078 and 2.722 visits.

C. About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.

Step-by-step explanation:

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 $s = \sqrt{\frac{p(1-p)}{n}}$

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

By the Central Limit Theorem, normal.

Question B:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.645.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645\frac{2.9}{\sqrt{219}} = 0.322$

The lower end of the interval is the sample mean subtracted by M. So it is 2.4 - 0.322 = 2.078 visits

The upper end of the interval is the sample mean added to M. So it is 2.4 + 0.322 = 2.722 visits.

Between 2.078 and 2.722 visits.

Question c:

90% confidence level, so 90% will contain the true population mean, 100 - 90 = 10% wont.

About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.

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3 years ago