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2 years ago

Complete the table. Will choose brainliest.

Mathematics

1 answer:

maxonik [38]2 years ago

5 0

Answer:

first one is 82% second is 214%

Step-by-step explanation:

Step 1: Divide 288 by 351 to get the number as a decimal. 288 / 351 = 0.82

Step 2: Multiply 0.82 by 100. 0.82 times 100 = 82

Step 1: Divide 212 by 99 to get the number as a decimal. 212 / 99 = 2.14

Step 2: Multiply 2.14 by 100. 2.14 times 100 = 214

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What is the slope simplify your answer and write it as a proper fraction, improper fraction, or integer

worty [1.4K]

Take 2 points: (0, 70) (30, 80)

SLOPE FORMULA: y2 -y1 / x2 - x1

y2 - y1: 80 - 70 = 10

x2 - x1: 30-0 = 30

10 / 30

simplify: 1/3

This is a PROPER FRACTION

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3 years ago

What is 1/3*4 I need it for homework I am working on Fraction Multiplication

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3 years ago

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aleksley [76]

Answer:

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3 years ago

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Dovator [93]

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3 years ago

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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

4 years ago