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mylen [45]
2 years ago
14

PLS HELP I NEED IT FOR MY TEST. I WILL GIVE BRAINLIEST.

Mathematics
1 answer:
CaHeK987 [17]2 years ago
8 0
Wouldn’t the number of bison decrease??
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Is this inequality statement true? (Y or N) <br><br> -13 &lt; -5
Kisachek [45]

Answer:

yes

Step-by-step explanation:

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3 years ago
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What is the solution to this equation x/-2 =18
svetlana [45]

x = - 36

multiply both sides of the equation by - 2

x = (- 2)× 18 = - 36



4 0
4 years ago
You recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the
il63 [147K]

Answer:

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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

Of the 2809 people who responded to survey, 1634 stated that they currently use social media.

This means that n = 2809, \pi = \frac{1634}{2809} = 0.5817

98% confidence level

So \alpha = 0.02, z is the value of Z that has a pvalue of 1 - \frac{0.02}{2} = 0.99, so Z = 2.327.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

6 0
3 years ago
2+8<br><br> PLSSS HELP PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
romanna [79]

Answer:

10

Step-by-step explanation

2 + 8= 10

6 0
3 years ago
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Simplify(4-x) (7+x) <br>​
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(4-x) ( 7+ ×)

28 +4x -7x -x²

28 -3x - x²

7 0
3 years ago
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