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2 years ago

What is the rate in which the second quantity is one unit

1 answer:

4 0

When grades are expressed as a quantity of 1, such as two feet per second or 5 miles per hour, they are called unit rates.

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Factor the expression 48g^2 -22gh -15h^2

sasho [114]

Answer:

factor the expression 48g^2 -22gh -15h^2

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3 years ago

Please look at this multiple choice, thanks!

Nastasia [14]

Step-by-step explanation:

$\frac{y - y1}{x - x1} = \frac{y2 - y1}{x2 - x1}$

$\frac{y - 0}{x - ( - 1)} = \frac{3 - 0}{1 - ( - 1)}$

$\frac{y}{x + 1} = \frac{3}{1 + 1}$

$\frac{y}{x + 1} = \frac{3}{2}$

$y = \frac{3}{2} (x + 1)$

$y = \frac{3}{2} x + \frac{3}{2}$

option B

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3 years ago

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3 years ago

What is the definition of a line

kolezko [41]

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3 years ago

Read 2 more answers

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

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

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

2 years ago