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2 years ago

The table displays the scores of students on a recent exam. Find the mean of the

Mathematics

1 answer:

Crank2 years ago

7 0

Answer:

please see the photo above for detailed analysis.

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Select the correct product (x2+4) (x2-4) x4 16 x4+16 x2+8x+16 x2 8x 16

natka813 [3]

The answer would be x^4 -16
Explanation: use the FOIL method.
(x^2 +4)(x^2 -4)= x^4 +4x -4x -16. 4x and -4x cancel each other out. You are left with x^4 -16.
F- irst
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8 0

3 years ago

I need help really bad

sdas [7]

Answer:

the x intercept=(-6,0) and y intercept=(0,14)

Step-by-step explanation:

i used math may but put the m as w, it works

5 0

2 years ago

How long will it take to burn 5 pounds playing rackuetball

Lady_Fox [76]

Answer:

120 hours

Step-by-step explanation:

24 hours in a day

playing a useless sport like that you only lose a single pound a day

so 5 *24 = 120 hours or 5 days'

8 0

3 years ago

Look at the equation 4x - 8.4 = 2x + 5. The terms 4x and 2x are like terms

xz_007 [3.2K]

In form ax + b = c: 2x - 8.4 = 5

Solution to equation: x = 6.7

Step-by-step explanation:

4x - 8.4 = 2x + 5

2x - 8.4 = 5

2x = 13.4

x = 6.7

3 0

3 years ago

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

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

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

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, 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 lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago