All categories

3 years ago

Which are partcial products for 34 x 28

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

3 0

Answer:

952

Step-by-step explanation:

34 = (30 + 4)

28 = (20 + 8)

34 x 28 = (30 + 4) x (20 + 8)

34 x 28 = 30 x 20 + 30 x 8 + 4 x 20 + 4 x 8

34 x 28 = 600 + 240 + 80 + 32

34 x 28 = 952

952 = 952

You might be interested in

Two numbers that multiply to 120 and add to 29

Veseljchak [2.6K]

10*12=120+29=149 hoped i helped

8 0

3 years ago

If m(x) = x2 + 3 and n(x) = 5x + 9, which expression is equivalent to (mn)(x)?

Romashka [77]

(x^2 +3)(5x+9)

5x^3 + 9x^2 + 15x + 27

3 0

3 years ago

In a clinical study, volunteers are tested for a gene that has been found to increase the risk for a disease. The probability th

MrRa [10]

Answer:

a) 0.984

b) 20 people

Step-by-step explanation:

If The probability that a person carries the gene is 0.1, then in a sample of 20 people, 2 should carry the gene.

Now, we want to know how many samples there are with this property.

Since we have 20 elements where 18 are alike (do not carry the gene) and 2 are alike (carry the gene), we have to compute the number of permutations of 20 elements in which 18 are alike and 2 are alike. This number is

$\frac{20!}{18!2!}=190$

In this 190 20-tuples there are only 3 where the 2 carriers of the gene are in the first 3 places, namely

(1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

where 1=carries the gene, 0=does not carry the gene.

So there are 190-3 = 187 elements in which the first 3 elements have no 2 carriers, hence the probability that 4 or more people will have to be tested until 2 of them with the gene are detected is 187/190 = 0.984 (98.4%) rounded to three decimal places.

Given that the probability that a person carries the gene is 0.1, then in a sample of 20 people, 20*0.1 = 2 should carry the gene.

4 0

3 years ago

Wait so the sum of interior angles

Amiraneli [1.4K]

<h3>Answer: 1260</h3>

====================================================

Work Shown:

n = number of sides = 9

S = sum of the interior angles of any polygon with n sides

S = 180(n-2)

S = 180(9-2)

S = 180(7)

S = 1260

8 0

3 years ago

A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence

Varvara68 [4.7K]

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

$n = 30$ -- sample size

$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {30}}$

$\sigma_x = \frac{10}{5.478}$

$\sigma_x = 1.825$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.825$

$E = 3.577$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 3.577$

$Lower = 46.423$

$Lower = 46.4$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 3.577$

$Upper = 53.577$

$Upper = 53.6$ --- approximated

<em>So, the 95% confidence interval is (46.4, 53.6)</em>

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

<em>So, the 95% confidence interval is (47.9, 52.1)</em>

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$ $E = 3.577$

In (b), we have:

$n = 90$ $E = 2.06584$

<em>Notice that the margin of error decreases when the sample size increases.</em>

4 0

3 years ago