Are these triangles similar? If so how? If not, explain.

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].

8 0

3 years ago

Mark the points A(5, 5) and B(7.2). Draw a circle that has point A as the center and passes through point B. Measure and label t

olasank [31]

Answer:

i think 4 units im not sure though though

Step-by-step explanation:

4 0

3 years ago

a study was to be undertaken to determine if a particular training program would improve physical fitness. an srs of 31 universi

Nataliya [291]

Step-by-step explanation:

A study was to be undertaken to determine if a particular training program would improve physical fitness. A sample of 31 university students was selected to be enrolled in the fitness program. The researchers wished to determine if there was evidence that their sample of students differed from the general population of untrained subjects. The sample mean is 47.4 and a standard deviation of 5.3. The 98% confidence interval is determined and is given as, (45.2, 49.6) .

If the level of confidence is changed to 95%, then the confidence interval will become shorter but the p-value will not change because it is calculated using the test statistic. So the correct answer is (a).

3 0

1 year ago

What is the value for x?

taurus [48]

Step-by-step explanation:

angle B=6x+4

6x=4

x=4/6

x=0.6

..I got this much...sorry if its wrong

5 0

3 years ago

Read 2 more answers

A survey of 200 families showed that

KengaRu [80]

Answer:

15 is the answer , the total people who has dogs , cats , parakeets and none is equals to <u>185</u> then from 200 you minus 185 which is 15 so 15 people had parakeets only

4 0

3 years ago