WILL MARK BRAINLIEST

5 of 8

aleksandr82 [10.1K]

Given:

The mean age of 5 people in a room is 26 years.

A person enters the room. The mean age is now 33.

To find:

The age of the person who entered the room.

Solution:

Formula for mean:

$\text{Mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}$

The mean age of 5 people in a room is 26 years.

$26=\dfrac{\text{Sum of ages of 5 people}}{5}$

$26\times 5=\text{Sum of ages of 5 people}$

$130=\text{Sum of ages of 5 people}$

The mean age is now 33. It means, the mean age of 6 people is 33.

$33=\dfrac{\text{Sum of ages of 6 people}}{6}$

$33\times 6=\text{Sum of ages of 6 people}$

$198=\text{Sum of ages of 6 people}$

Now, the age of the person who entered the room is

Required age = Sum of ages of 6 people - Sum of ages of 5 people

= $198-130$

= $68$

Therefore, the age of the person who entered the room is 68 years.

6 0

3 years ago

Add the two expressions.<br> 3x + 7 and 92 - 2<br> Enter your answer in the box.<br> Help

Paul [167]

The answer to the question is 3x+97

3 0

2 years ago

There are 144 children performing in 3 school bands each band has the same number of children how many children are performing i

Readme [11.4K]

Answer:

48 children performing.

Step-by-step explanation:

Well all you do is divide 144 by 3 and you will get the answer of 48.

3 0

3 years ago

Read 2 more answers

A store pays $840 for an oil painting. The store marks up the price by 46%. What is the amount of the mark-up? please help!!!! I

Setler [38]

Answer:

A store pays $840 for an oil painting. The store marks up the price by 46%. What is the amount of the mark-up?

=

$386.40

3 0

2 years ago

Several years ago, 50% of parents who had children in grades K-12 were satisfied with the quality of education the students r

galina1969 [7]

Answer:

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

We have to see if 50% = 0.5 is part of the confidence interval.

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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 1095, \pi = \frac{478}{1095} = 0.4365$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 - 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4118$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 + 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4612$

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.

5 0

2 years ago