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r-ruslan [8.4K]
3 years ago
13

!LOOK AT THE PICTURE!Simplify 15.6 divided by negative 3

Mathematics
1 answer:
muminat3 years ago
3 0

Answer:

c -5.2 tell me if you need the explanation

Step-by-step explanation:

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Please answer correctly !!!!! Will mark Brianliest !!!!!!!!!!!!!!
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Answer:

1017.9ft ^{3}

Step-by-step explanation:

volume \: of \: cylinder \\ v = \pi \times r {}^{2}  \times h \\  = \pi  \times 9^2 \times 4 \\  = 1017.9

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Use the distributive property to write an equivalent expression.<br> 9(8w + 4)
Harlamova29_29 [7]

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72w+36

Step-by-step explanation:

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Step-by-step explanation:

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3 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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