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2 years ago

What is the equation in slope-intercept form of a line that has a slope of -- and a y-intercept of 6?

Mathematics

1 answer:

Elza [17]2 years ago

5 0

Answer:

Step-by-step explanation:

I fits the graph that is described

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Find angles in isosceles triangles

Tanzania [10]

Answer - I think the answer is 59

Step-by-step explanation:

6 0

2 years ago

Solve. <br> (2/3) (1/4) =

stellarik [79]

You multiply the numerators (top numbers) and separately multiply the (bottom numbers) denominators.

numerator (top number): 2 * 1 = 2

denominator (bottom number): 3 * 4 = 12

Now, we have 2/12 which can be simplified to 1/6.

6 0

3 years ago

How can you rewrite the expression (8-5i)^2 in the form a+bi?

otez555 [7]

(a-b)^2 = a^2-2ab+b^2

(8-5i)^2 = 8^2-2(8)(5i)+(5i)^2

= 64-80i+25i^2

i^2=-1

So

= 64-80i+25(-1)

=64-25-80i

= <em><u>39 - 80i</u></em>

which is your answer :)

8 0

3 years ago

Read 2 more answers

If the ratio of the length of a rectangle to its width is 9/4 and its length is 18 cm, what is the width of the rectangle?

Kruka [31]

Answer:

The width is 8

Step-by-step explanation:

if the ratio of length to width is 9 to 4. and we know the length is 18, 18 is 2 times as much as 9, so 4*2=8. 9*2=18,4*2=8

6 0

2 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