All categories

3 years ago

On his 8 tests, Hong earned a total of 568 points. What was his mean number if points earned per test?

Mathematics

1 answer:

madam [21]3 years ago

6 0

Answer:

He earned 71 points per test on average.

Step-by-step explanation:

Divide the total amount of points by the amount of tests to find the average (mean.)

You might be interested in

TIMED TEST!!! PLZ HELP! WILL GIVE BRAINLIEST!!!

aleksandrvk [35]

But but I thought asian where good at math

4 0

3 years ago

Read 2 more answers

Find the perimeter of each figure 8 ft (square)

sergey [27]

Because this figure is a square, multiply 8 by the number of sides the polygon has
in this case, it has 4 sides so multiply 8 by 4
8•4=32

4 0

3 years ago

A single story house is to be built on a rectangular lot 70 feet wide by 100 feet deep. The shorter side of the lot is along the

dedylja [7]

Answer:

250 sq ft

Step-by-step explanation:

Since the shorter side is along the street and the setback is 10ft from each side, the house is 50 ft wide

and 100 -20 -30 = 50 ft long

50 x 50 = 250 sq ft

3 0

4 years ago

A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute

GREYUIT [131]

Answer:

(a) Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute

Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute

(b) The value of chi-square test statistics is 35.704.

(c) P-value = 0.4360.

(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute.

Let $\sigma$ = population standard deviation for the pulse rates of men.

(a) So, Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is One-sample chi-square test for standard deviation;

T.S. = $\frac{(n-1)\times s^{2} }{\sigma^{2} }$ ~ $\chi^{2}__n_-_1$

where, s = sample standard deviation = 10.1 beats per minute

n = sample of men = 36

So, the test statistics = $\frac{(36-1)\times 10.1^{2} }{10^{2} }$ ~ $\chi^{2}__3_5$

= 35.70 4

(b) The value of chi-square test statistics is 35.704.

(c) Also, the P-value of the test statistics is given by;

P-value = P( $\chi^{2}__3_5$ > 35.704) = 0.4360

(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

3 0

3 years ago

40 POINTS Which expression is equivalent to (3k - 4b) ^ 3 in expanded form?

GuDViN [60]

The Answer is B

I just took the test

3 0

3 years ago