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3 years ago

What is the sum of the series?

Mathematics

1 answer:

otez555 [7]3 years ago

4 0

Okay, the number above the sigma(4) indicates what value you go to, and what's below the sigma tells you where to start, and the stuff on the right is the expression you're using.

So we start with 1 and go to 4, adding all the values in between.

(2(1))^2 + (2(2))^2 + (2(3))^2 + (2(4))^2

2^2 + 4^2 + 6^2 + 8^2

4 + 16 + 36 + 64

120

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Jared painted 5/12 and 4/12 blue.what part of the mug is not red or blue?

Reika [66]

3/12 of the mug is not blue or red

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3 years ago

Use the diagram to determine which statement is true

Dahasolnce [82]

The answer is d.

Finding area of ABCD :

Find side length

side = √3² + 4²
side = 5

2. Apply formula to find area

area = 5²
area = 25

Finding area of GHIA :

area = 4²
area - 16

Finding area of DEFG :

area = 3²
area = 9

Now, let's see whether is true.

Area (ABCD) - Area (GHIA) = Area (DEFG)
25 - 16 = 9
9 = 9

∴ Hence, it is proved √

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1 year ago

Please give the answer

dybincka [34]

Answer:

When N is an even number, use the first equation in the matrix. When N is negative use the second equation.

Step-by-step explanation:

n = 6 | 6/2 = 3

n=5 | 3(5)+1 = 16

5 0

2 years ago

Suppose that 6 out of 24 people leave a meeting early. what percent of the people leave early

Alisiya [41]

25% of people leave early as 6 is 1/4 of 24. You can solve this by dividing 6 by 24 (I believe, maybe check with someone on that but that’s how i figured it out).

4 0

3 years ago

Read 2 more answers

10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the

Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

$H_0: \mu = 0$

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

$H_1: \mu > 0$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

s is the standard deviation of the sample.

n is the sample size.

In this problem, the parameters are given as follows:

$\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50$ .

Hence, the test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}$

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by $t^{\ast} = 2.4$ .

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

5 0

1 year ago