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3 years ago

A/b + c/b=? Simmmmmpppppllllliiiifffyyy

Mathematics

2 answers:

Rina8888 [55]3 years ago

4 0

Answer: i think is like this

a+b

——

liubo4ka [24]3 years ago

3 0

Answer:

\frac{A+c}{b}

Step-by-step explanation:

\frac{A}{b}+\frac{c}{b}

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What is the sum of the first 9 terms of 7+21+63+...

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Step-by-step explanation:

+63 =91

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The product of the square of a number and 5

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Let the number be x, the square of x is x^2, then
The product of the square of x and 5 is 5x^2.

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How many solutions does 6x^2+40=-31x have?

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3 years ago

Dose anyone know how to solve this?

Reil [10]

Answer:

3x³-6x²
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Step-by-step explanation:

For 1: Multiply both functions together, expand, then simplify

(x-2)(3x²) = 3x³-6x²

For 2: Just add both functions together

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2 years ago

Read 2 more answers

You are given the following information obtained from a random sample of 5 observations. 20 18 17 22 18 At 90% confidence, you w

Margaret [11]

Answer:

The null hypothesis is

$H_o : \mu = 21$

The Alternative hypothesis is

$H_a : \mu< 21$

$\sigma_{\= x} = 0.8944$

$t = -2.236$

Yes the mean population is significantly less than 21.

Step-by-step explanation:

From the question we are given

a set of data

20 18 17 22 18

The confidence level is 90%

The sample size is n = 5

Generally the mean of the sample is mathematically evaluated as

$\= x = \frac{20 + 18 + 17 + 22 + 18}{5}$

$\= x = 19$

The standard deviation is evaluated as

$\sigma = \sqrt{ \frac{\sum (x_i - \= x)^2}{n} }$

$\sigma = \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }$

$\sigma = 2$

Now the confidence level is given as 90 % hence the level of significance can be evaluated as

$\alpha = 100 - 90$

$\alpha = 10$ %

$\alpha =0.10$

Now the null hypothesis is

$H_o : \mu = 21$

the Alternative hypothesis is

$H_a : \mu< 21$

The standard error of mean is mathematically evaluated as

$\sigma_{\= x} = \frac{\sigma}{ \sqrt{n} }$

substituting values

$\sigma_{\= x} = \frac{2}{ \sqrt{5 } }$

$\sigma_{\= x} = 0.8944$

The test statistic is evaluated as

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

substituting values

$t = \frac{ 19 - 21 }{ 0.8944 }$

$t = -2.236$

The critical value of the level of significance is obtained from the critical value table for z values as

$z_{0.10} = 1.28$

Looking at the obtained value we see that $z_{0.10}$ is greater than the test statistics value so the null hypothesis is rejected

6 0

3 years ago