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sp2606 [1]
3 years ago
14

The area of a square is 9 centimeters squared (cm^2) If the sides of the square are doubled in length, what is the new area, in

centimeters squared, of the enlarged square?
A: 24 cm^2
B: 18 cm^2
C: 81 cm^2
D: 36 cm^2
Mathematics
2 answers:
Mice21 [21]3 years ago
7 0

Answer:

The question says area of a square = 9 cm squared.

And they said if the sides are doubled in length what is the new length

so new length = 9 cm^2 × 9 cm^2

which should give you 81cm^2

therefore your answer should be = C

Sauron [17]3 years ago
4 0
I think the answer is B because 9 cm times 2 cm equal 18
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Step-by-step explanation:

Since we're rounding to the hundredths place, we need to see if the thousandths place is greater or equal to 5. Since it is less than 5, we round down to 72.63.

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Andru [333]

Answer:

The 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population standard deviation is:

CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}

The information provided is:

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\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.10/2, (26-1)}=\chi^{2}_{0.05, 25}=37.652

\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.10/2, (26-1)}=\chi^{2}_{0.95, 25}=14.611

*Use a Chi-square table.

Compute the 90% confidence interval for the population standard deviation waiting time for an oil change as follows:

CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}

     =\sqrt{\frac{(26-1)\times 4.8^{2}}{37.652}}\leq \sigma\leq \sqrt{\frac{(26-1)\times 4.8^{2}}{14.611}}\\\\=3.9113\leq \sigma\leq 6.2787\\\\\approx 3.9 \leq \sigma\leq6.3

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Answer:

<u>Therefore, the value of x is 5.</u>

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<u />

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