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2 years ago

Find the length of the second leg of the triangle round to the nearest hundredth

Mathematics

1 answer:

horsena [70]2 years ago

7 0

Answer:

$\displaystyle \frac{3\sqrt{230}}{5}$

Step-by-step explanation:

Use the Pythagorean Theorem:

$\displaystyle a^2 + b^2 = c^2 \\ \\ 9,7^2 + b^2 = 13,3^2; \sqrt{82,8} = \sqrt{b^2} \\ \\ \frac{3\sqrt{230}}{5}\:[or\:9,0994505329...] = b$

So, you have this:

$\displaystyle 9,1 ≈ b$

I am joyous to assist you at any time.

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Which relation is also a function? A, B, C, or D?

andrew-mc [135]

Answer:

AorD

srry if im wrong its not B because some of them are not compatible

Step-by-step explanation:

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2 years ago

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A survey of 250 homeless persons showed that 14 were veterans. Find a point estimate for the population proportion of homeless p

snow_lady [41]

Answer:

$\hat p =\frac{14}{250}= 0.056$

And for this case that represent the 5.6% of the homeless persons who are veterans

Step-by-step explanation:

We know that in a survey of 250 homeless persons 14 were veterans. We can find the estimated proportion of homeless persons who are veterans with the following formula:

$\hat p =\frac{X}{n}$

The last one is an unbiased estimator of the true population proportion p

And replacing the info given we got:

$\hat p =\frac{14}{250}= 0.056$

And for this case that represent the 5.6% of the homeless persons who are veterans

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2 years ago

How can you write an equation for a line from one point and the slope?

Andrew [12]

Answer y=mx+b

Step-by-step explanation: b is the y intercept and x is the slope

5 0

2 years ago

Consider the following equation, bh+hr=25 when solving for r

Komok [63]

Solve for r.

You want to get r by itself on one side on the equal sign.

bh + hr = 25

Subtract bh from both sides.

hr = 25 - bh

Divide h on both sides.

r = 25 - bh / h

The two h's cancel each other out.

r = 25 - b

Hope this helps!

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2 years ago

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The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$