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motikmotik
3 years ago
12

Solve for x. pls help !!!!!

Mathematics
1 answer:
yanalaym [24]3 years ago
5 0

Answer:

x=5

Step-by-step explanation:

Put everything on one side of the equation and have it equal to 180 degrees.

13x+3+72+8x= 180

Combine like terms

21x+75= 180

Subtract 75 on both sides

21x = 105

Divide by 21 on both sides

x= 5

hope this helped!

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Julio wants to break his school’s scoring record of 864 points during his 24-game basketball season. During the first 8 games of
Alecsey [184]

Answer:

Hello There the Correct answer is 8 + 24 x greater-than 864.

<h3> Explanation</h3>

Step 1: 864-56= 608

Step 2: 24-8= 16

Step 3: 608/16=38

7 0
4 years ago
Read 2 more answers
Juries should have the same racial distribution as the surrounding communities. According to the U.S. Census Bureau, 18% of resi
Ymorist [56]

Answer:

0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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}}

18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American.

This means that p = 0.18, n = 100

So, by the Central Limit Theorem:

\mu = 0.18, s = \sqrt{\frac{0.18*0.82}{100}} = 0.0384

How likely is the resulting sample proportion to be between 0.066 and 0.294?

This is the pvalue of Z when X = 0.294 subtracted by the pvalue of Z when X = 0.066. So

X = 0.294

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.294 - 0.18}{0.0384}

Z = 2.97

Z = 2.97 has a pvalue of 0.9985

X = 0.066

Z = \frac{X - \mu}{s}

Z = \frac{0.066 - 0.18}{0.0384}

Z = -2.97

Z = -2.97 has a pvalue of 0.0015

0.9985 - 0.0015 = 0.997

0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294

3 0
3 years ago
if you subtract 17 from my number and multiply the difference by -6 the results is -138 what is Sarah's number
likoan [24]

Answer:

40

Step-by-step explanation:

let the unknown be y (x is used as times(multiplication) )

(y-17)x-6=-138

add /-6 to both sides

y-17x-6/-6=-138/-6

y-17=23

make y independent/add 17 to both sides

y-17+17=23+17

y=40

or in an easier way

-138/-6 =23

23+17=40

simply reversing the question to get the answer

7 0
3 years ago
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dedylja [7]

Answer:

7<em>x</em> + 10

Step-by-step explanation:

10 + 8<em>x</em> - 1<em>x</em> ​

<em>Add or subtract like terms.</em>

10 + (8 - 1)<em>x</em>

10 + (7)<em>x</em>

<em>Rearrange.</em>

7<em>x</em> + 10

7 0
3 years ago
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Simplify the expressions:
mestny [16]

Answer:

-1v

0.6y+2.1

3w-p+4

1-6x

0

-5x+10y+z-25

Step-by-step explanation:

Combine like terms

8 0
3 years ago
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