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3 years ago

Simplify:

Mathematics

1 answer:

wlad13 [49]3 years ago

5 0

Answer:

The answer to your question is letter C. 7y³ + 7n²y² - 22y²

Step-by-step explanation:

y²(4y + 7n² + 2) - 3y² (-y + 8)

Multiply

4y³ + 7n²y² + 2y² + 3y³ - 24y²

Use the associative property for like terms

(4y³ + 3y³) + (2y² - 24y²) + 7n²y²

Simplify like terms

7y³ - 22y² + 7n²y² or 7y³ + 7n²y² - 22y²

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Which of the following equations is of a parabola with a vertex at (0, -5)

Harlamova29_29 [7]

The x, which is zero in this case will always be the number that goes on the inside of the equation, so far you have y=(x+0)^2 or y=x^2.

The y, which is negative will go on the end of the equation. So you get y=x^2-5

The equation is y=x^2-5 or C if it is multiple choice

4 0

3 years ago

Read 2 more answers

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[

Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

$= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

$= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}\\\\$

calculate the sum: $S= s_1+s_2+s_3+s_n\\$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$

$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$

8 0

3 years ago

PLZZZZZZZZZZZZZZZZZ HELP!

zheka24 [161]

Answer:

-8

Step-by-step explanation:

5(x+12)= 20

5x+60= 20

5x= -40

x= -8

3 0

3 years ago

The mean per capita income is 16,127 dollars per annum with a variance of 682,276. What is the probability that the sample mean

MakcuM [25]

Answer:

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The standard deviation is the square root of the variance. So

$\mu = 16127, \sigma = \sqrt{682276} = 826, n = 476, s = \frac{826}{\sqrt{476}} = 37.86$

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Either it differs by 104 or less dollars, or it differs by more than 104 dollars. The sum of the probabilities of these events is 100. I am going to find the probability that it differs by 104 or less dollars first.

Probability that it differs by 104 or less dollars first.

pvalue of Z when X = 16127 + 104 = 16231 subtracted by the pvalue of Z when X = 16127 - 104 = 16023. So

X = 16231

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{16231 - 16127}{37.86}$

$Z = 2.75$

$Z = 2.75$ has a pvalue of 0.9970

X = 16023

$Z = \frac{X - \mu}{s}$

$Z = \frac{16023 - 16127}{37.86}$

$Z = -2.75$

$Z = -2.75$ has a pvalue of 0.0030

0.9970 - 0.0030 = 0.9940

99.40% probability that it differs by 104 or less.

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

p + 99.40 = 100

p = 0.60

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

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3 years ago

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Brilliant_brown [7]

$\boxed{ \tt{its \: a \: straight \: line \: graph}}$

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