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3 years ago

PLEASE HELP WITH 10 AND 11

Mathematics

1 answer:

Gnoma [55]3 years ago

6 0

-2,8 5,3
slope is y2--y1 / x2-x1

3-8 -5
___ ____
5- -2 7

10 is -5/7
11 is 0

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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

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3 years ago

-6d=-42 can you help me solve please

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Given the equation :

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So, to find d , divide both sides by ( -6 )

So,

$\begin{gathered} \frac{-6d}{-6}=\frac{-42}{-6} \\ \\ d=7 \end{gathered}$

Note: when we divide two negative numbers, the result will be a positive number

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Step-by-step explanation:

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