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Arturiano [62]
1 year ago
5

Which of the following Roots would be between 8 and 7

Mathematics
1 answer:
azamat1 year ago
3 0

To find which of the following roots is between "8" and "7" we can calculate the root of which numbers result in 8 and 7. To do this we will power them by 2, this is done because power is the oposite operation to the root. Doing this gives us:

\begin{gathered} 8^2=64 \\ 7^2=49 \end{gathered}

So the root of 64 is 8 and the root of 49 is 7. We need to find the number that is between 49 and 64.

From the options the only one that qualifies is 52. The correct option is b.

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enyata [817]
Umm what? I don’t think i can answer this cause i can barely read it.
8 0
4 years ago
[3x-7]=12 solve for x
kompoz [17]
3x - 7 + 7 = 12 + 7
3x = 19
3x/3 = 19/3
X = 19/3 = 6 1/3
8 0
3 years ago
Read 2 more answers
-2 1/4 - ( - 3/4) =<br>​
Grace [21]

Answer: -1.5

Step-by-step explanation:

6 0
3 years ago
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If you answer this can you explain how you got that answer please and thank you
liberstina [14]

Answer:

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Step-by-step explanation:

5 0
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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