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Ostrovityanka [42]
4 years ago
7

Point x is located at (-3,-4) and point Y is located at (6,-4). What is the distance between these two points?

Mathematics
1 answer:
mart [117]4 years ago
7 0

Answer:

9

Step-by-step explanation:

Distance between two points is given as;

Square root [(X2-x1) 2+ (y2-y1)2]

= (6-(-3))2 +(-4-(-4))2

=Square root (81 +0)

=9

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GarryVolchara [31]

The digit in the hundred millions place is 9.

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What is the slope that passes through the points : (-2,9) and (4,-7)
Westkost [7]

Use the slope formula below:

\large \boxed{m =  \frac{y_2 - y_1}{x_2 - x_1} }

The m-term represents the slope.

The formula is the changes of two y-points over the changes of two x-points. We are given two points. Substitute both points in the formula.

\large{m =  \frac{9 - ( - 7)}{ - 2 - 4} } \\  \large{m =  \frac{9 + 7}{ - 6}  \longrightarrow  \frac{16}{ - 6} } \\  \large{m =  \frac{8}{ -3 }  \longrightarrow  -  \frac{8}{3} }

Therefore the slope is -8/3

Answer

  • the slope is -8/3

Hope this helps! Let me know if you have any doubts.

6 0
3 years ago
Read 2 more answers
s defined to be the dollar value of loans defaulted divided by the total dollar value of all loans made. Banking officials claim
Amiraneli [1.4K]

Answer:

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

Test statistic t = 1.431

Critical values tc = ±2.447

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Step-by-step explanation:

This is a hypothesis t-test for the population mean.

The claim is that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

Then, the null and alternative hypothesis are:

H_0: \mu=3.5\\\\H_a:\mu\neq 3.5

The significance level is 0.05.

The sample has a size n=7.

We calculate  the sample mean and standard deviation as:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{7}(7+4+6+3+5+4+2)\\\\\\M=\dfrac{31}{7}\\\\\\M=4.43\\\\\\

s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{6}((7-4.43)^2+(4-4.43)^2+(6-4.43)^2+. . . +(2-4.43)^2)}\\\\\\s=\sqrt{\dfrac{17.71}{6}}\\\\\\s=\sqrt{2.95}=1.72\\\\\\

The sample mean is M=4.43.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.72.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1.72}{\sqrt{7}}=0.65

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t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{4.43-3.5}{0.65}=\dfrac{0.93}{0.65}=1.431

The degrees of freedom for this sample size are:

df=n-1=7-1=6

This test is a two-tailed test, with 6 degrees of freedom and t=1.431, so the P-value for this test is calculated as (using a t-table):

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If we use the critical value approach, for this level of confidence, the critical values are tc = ±2.447. The test statistic is within the bounds of the critical values and falls within the acceptance region.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

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