Can someone please be amazing and help me solve number 2?

The average of a list of 4 numbers is 90.0. A new list of 4 numbers has the same first 3 numbers as the original list, but the f

Thepotemich [5.8K]

Answer:

the average of this new list of numbers is 94

Step-by-step explanation:

Hello!

To answer this question we will assign a letter to each number for the first list and the second list of numbers, remembering that the last number of the first list is 80 and the last number of the second list is 96

for the first list

$\frac{a+b+c+80}{4} =90$

for the new list

$\frac{a+b+c+96}{4} =X$

To solve this problem consider the following

1.X is the average value of the second list

2. We will assign a Y value to the sum of the numbers a, b, c.

a + b + c = Y to create two new equations

for the first list

$\frac{y+80}{4} =90$

solving for Y

Y=(90)(4)-80=280

Y=280=a+b+c

for the second list

$\frac{y+96}{4} =X\\$

$\frac{280+96}{4} =X\\x=94$

the average of this new list of numbers is 94

4 0

3 years ago

Read 2 more answers

Rewa Delta Union Rugby CEO has become concerned about the slow pace of the rugby games played in the current union rugby, fearin

ozzi

Answer:

We conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

Step-by-step explanation:

We are given that Before the meeting, the mean duration of the 15-sided rugby game time was 3 hours, 5 minutes, that is, 185 minutes.

A random sample of 36 of the 15-sided rugby games after the meeting showed a mean of 179 minutes with a standard deviation of 12 minutes.

Let $\mu$ = mean duration of 15-sided union rugby games after the meeting.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 185 minutes {means that the mean duration of 15-sided union rugby games has increased or remained same after the meeting}

Alternate Hypothesis, $H_A$ : $\mu$ < 185 minutes {means that the mean duration of 15-sided union rugby games has decreased after the meeting}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean duration of 15-sided union rugby games = 179 min

s = sample standard deviation = 12 minutes

n = sample of 15-sided rugby games = 36

So, the test statistics = $\frac{179-185}{\frac{12}{\sqrt{36} } }$ ~ $t_3_5$

= -3

The value of t test statistics is -3.

Now, at 1% significance level the t table gives critical value of -2.437 at 35 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of t as -3 < -2.437, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

6 0

4 years ago

Geometry peculiar pool

saw5 [17]

Answer:

hi.

Step-by-step explanation:

download photo math

5 0

3 years ago

(PLEASE ANSWER ASAP) Which number is between 1/6 and 0.25?

serg [7]

If you change both to a fraction you would see that you have 1/6 and 1/4 (.25 as a fraction). So 1/5 would be in between them.

You could also change both as a decimal. 1/6 =.166666 and .25 so any number in between them would be correct such as .2 which is the same as 1/5.

8 0

4 years ago

Jessica is deciding on her schedule for next semester. She must take each of the following classes: English 101, Spanish 102, Bi

user100 [1]

Just a random guess 22,275 tell me if it right

8 0

3 years ago