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2 years ago

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calculates the sum of the first 8 terms of an arithmetic progression starting with 3/5 and ending with 1/4

1 answer:

N76 [4]2 years ago

6 0

We conclude that the sum of the first 8 terms of the arithmetic sequence is 17/5.

<h3>How to get the sum of the first 8 terms?</h3>

In an arithmetic sequence, the difference between any two consecutive terms is a constant.

Here we know that:

$a_1 = 3/5\\a_8 = 1/4$

There are 7 times the common difference between these two values, so if d is the common difference:

$a_1 + 7*d = a_8\\\\3/5 + 7*d = 1/4\\\\7*d = 1/4 - 3/5 = (5 - 12)/20 = -7/20\\\\d = -1/20$

Then the sum of the first 8 terms is given by:

$3/5 + (3/5 - 1/20) + (3/5 - 2/20) + ... + (3/5 - 7/20)\\\\8*(3/5) - (1/20)*(1 + 2 + 3+ 4 + 5 + 6 + 7) = 3.4 = 34/10 = 17/5$

So we conclude that the sum of the first 8 terms of the arithmetic sequence is 17/5.

If you want to learn more about arithmetic sequences:

brainly.com/question/6561461

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Step-by-step explanation:

30 to 65 is 35 years

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4 years ago

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3 years ago

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3 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago

. Suppose you plan to make a box that will hold exactly 40 one-inch cubes. Give the dimensions (ex: 1in x 1in x 40in) of 3 possi

stiks02 [169]

Answer:

The dimensions of the three possible boxes are;

Box 1: 2 inches length by 4 inches width by 5 inches height

Box 2: 2 inches length by 2 inches width by 10 inches height

Box 3: 8 inches length by 1 inch width by 5 inches height

Step-by-step explanation:

The dimensions of the required box = 40 in³

Therefore, a box that will hold all the cubes will have a dimension given as follows;

Length, l × Width, w × Height, h = 40 in.³

We therefore, look for possible dimensions of l, w, h, however to make is faster, we note that l × w = base area, BA, hence we look for the factors 40 thus;

BA × h ≡ l × w × h

8 × 5 = 40 ≡ 2 × 4 × 5 = 40

4 × 10 = 40 ≡ 2 × 2 × 10 = 40

8 × 5 = 40 ≡ 8 × 1 × 5 = 40

Hence we have our three possible boxes as follows;

Box 1: Length = 2 inches, width = 4 inches and height = 5 inches

Box 2: Length = 2 inches, width = 2 inches and height = 10 inches

Box 3: Length = 8 inches, width = 1 inches and height = 5 inches.

4 0

3 years ago