Somebody please help with correct answer and explanation??!!

Mathematics

1 answer:

kati45 [8]3 years ago

5 0

Hi there! The answer is A. (0, a).

When we plug in the x-coordinate we get:
$f(0) = a \times {b}^{0} = a \times 1 = a$

When we fill in the x-coordinate of any other point we don't get a proper outcome.

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Goshia [24]

Answer:

Step-by-step explanation:

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3 years ago

Mr. Good Wrench advertises that a customer will have to wait no more than 30 minutes for an oil change. A sample of 26 oil chang

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 - α)% 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:

n = 26

s = 4.8 minutes

Confidence level = 90%

Compute the critical values of Chi-square as follows:

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

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

*Use a Chi-square table.

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

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

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