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quester [9]
3 years ago
9

Find the slope of the line through each pair of points. (17,-6),(-11,7)

Mathematics
1 answer:
jeka57 [31]3 years ago
8 0
Slope is rise over run or
(y1-y2)/(x1-x2)

(x,y)
x1=17
x2=-11
y1=-6
y2=7

slope=(-6-7)/(17-(-11))=(-13)/(17+11)=(-13)/(28)=

the slope is -13/28
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How do you write 3.625 as a fraction
vesna_86 [32]
3.625 as a fraction would be 3 5/8 or 29/8.
You would get that by converting 3.625 to 3 625/1000. You do that because there is a 3 before the decimal- (that is a whole #) and three digits after the decimal (that means it goes into the thousandths place, so the denominator is 1000), Then you take the numbers after the decimal and place it over the 1000. Then you would simply the 3 625/1000 by 125. and get 3 5/8 or as a mixed # 29/8. (Multiply the 8 and 3 to get 24 and add the numerator (5) to get 29. The denominator stays the same.

Hope this helps!
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3 years ago
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Sergio039 [100]

Answer:

Step-by-step explanation:

4 0
2 years ago
Some transportation experts claim that it is the variability of speeds, rather than the level of speeds, that is a critical fact
scZoUnD [109]

Answer:

Explained below.

Step-by-step explanation:

The claim made by an expert is that driving conditions are dangerous if the variance of speeds exceeds 75 (mph)².

(1)

The hypothesis for both the test can be defined as:

<em>H</em>₀: The variance of speeds does not exceeds 75 (mph)², i.e. <em>σ</em>² ≤ 75.

<em>Hₐ</em>: The variance of speeds exceeds 75 (mph)², i.e. <em>σ</em>² > 75.

(2)

A Chi-square test will be used to perform the test.

The significance level of the test is, <em>α</em> = 0.05.

The degrees of freedom of the test is,

df = n - 1 = 55 - 1 = 54

Compute the critical value as follows:

\chi^{2}_{\alpha, (n-1)}=\chi^{2}_{0.05, 54}=72.153

Decision rule:

If the test statistic value is more than the critical value then the null hypothesis will be rejected and vice-versa.

(3)

Compute the test statistic as follows:

\chi^{2}=\frac{(n-1)\times s^{2}}{\sigma^{2}}

    =\frac{(55-1)\times 94.7}{75}\\\\=68.184

The test statistic value is, 68.184.

Decision:

cal.\chi^{2}=68.184

The null hypothesis will not be rejected at 5% level of significance.

Conclusion:

The variance of speeds does not exceeds 75 (mph)². Thus, concluding that driving conditions are not dangerous on this highway.

7 0
3 years ago
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liq [111]

Answer:

Equation : y = -0.5x − 8

Step-by-step explanation:

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