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soldier1979 [14.2K]
3 years ago
5

Apply the distributive property to create an equivalent expression. 1/2 (2a-6b+8)=

Mathematics
1 answer:
Rus_ich [418]3 years ago
3 0

Answer:

a-3b+4

Step-by-step explanation:

a-3b+4

just divide 2a-6b+8 with 2

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Please someone help me with this ASAP especially with the workout!!
lara31 [8.8K]

Answer:

71 degrees

Step-by-step explanation:

please someone help me with this ASAP especially with the workout!!

From the triangle shown;

YX² = YZ² + XZ² - 2(YZ)(XZ) cos Z

28² = 25² + 23² - 2(25)(23) cos Z

784 = 625 + 529- 1150 cos Z

784 - 1154 = -1150cos Z

-370 = -1150cos Z

cos Z = 370/1150

cos Z = 37/115

cos Z = 0.3217

Z = arccos 0.3217

Z = 71.23

hence the measure of <Z is 71 degrees

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3 years ago
(-19, -12), (-9, -15)
Jlenok [28]

Answer:

-28/-27 is the answer to find slope

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3 years ago
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stepan [7]
Did you not learn to factor?
6 0
3 years ago
The producer of a certain bottling equipment claims that the variance of all its filled bottles is .027 or less. A sample of 30
o-na [289]

Answer:

p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459

b. between .025 and .05

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

\bar X represent the sample mean

n = 30 sample size

s= 0.2 represent the sample deviation

\sigma_o =\sqrt{0.027}=0.164 the value that we want to test

p_v represent the p value for the test

t represent the statistic

\alpha= significance level

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is less than 0.027, so the system of hypothesis are:

H0: \sigma \leq 0.027

H1: \sigma >0.027

In order to check the hypothesis we need to calculate the statistic given by the following formula:

t=(n-1) [\frac{s}{\sigma_o}]^2

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

t=(30-1) [\frac{0.2}{0.164}]^2 =42.963

What is the approximate p-value of the test?

The degrees of freedom are given by:

df=n-1= 30-1=29

For this case since we have a right tailed test the p value is given by:

p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459

And the best option would be:

b. between .025 and .05

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3 years ago
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