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algol13
3 years ago
11

I need help on this plsssssssssssssssssssssssss

Mathematics
1 answer:
Savatey [412]3 years ago
4 0

Answer:

okay the answer is d.

Step-by-step explanation:

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Month 1 2 3 4 5 6 7
ikadub [295]

Answer:

Following are the solution to this question:

Step-by-step explanation:

Please find the complete question and its solution file in the attachment.

6 0
3 years ago
According to one web page on the resale value of cars, a BMW that cost $71,000 in 2003 depreciates in value about 15% a year. Ho
dedylja [7]
Depreciate- to lose value over time

it loses 15% over one year meaning that it retains 85% of its value from the previous year<span>.

2003-2013= 10 years

71,000 * (0</span>.85^10yrs)= $13,978<span>.</span>08

Do you have any questions?
8 0
3 years ago
Simplify the answer
yanalaym [24]

Answer:

{a }^{2}

this is the answer

3 0
3 years ago
What is the radius and diameter of the following circle?
Cerrena [4.2K]

Answer:

Radius=9

Diameter=18

Step-by-step explanation:

5 0
3 years ago
A random sample of 60 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact te
pentagon [3]

Answer:

The 95% confidence interval on the true proportion of helmets of this type that would show damage from this test is (0.169, 0.397).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 60, \pi = \frac{17}{60} = 0.283

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.283 - 1.96\sqrt{\frac{0.283*0.717}{60}} = 0.169

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.283 + 1.96\sqrt{\frac{0.283*0.717}{60}} = 0.397

The 95% confidence interval on the true proportion of helmets of this type that would show damage from this test is (0.169, 0.397).

6 0
3 years ago
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