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SpyIntel [72]
3 years ago
10

Simplify.

Mathematics
2 answers:
Sonbull [250]3 years ago
8 0

Answer:

a

Step-by-step explanation:

NikAS [45]3 years ago
4 0

= (16x3 - 8x2 + 4x4) / 2x

= 4x * (4x2 - 2x + x3) / 2x

= 2 * (4x2 - 2x + x3)

= 8x2 - 4x + 2x3

Answer A

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Find the range of the function. Use interval notation. g(x)=log↓3(x-2)​
Marizza181 [45]

Answer:

10x=log6-3

Step-by-step explanation:

5 0
2 years ago
Football team play $28 per jersey. They bought 16 jerseys. How much money did the team spend on jerseys
yaroslaw [1]
If they pay $28 for 16 jerseys then the total they’d have to play is $448

28 x 16 = 448

Hope this helps :)
5 0
3 years ago
it snowed 1/4 inch every hour for 4 hours then it's no 1/2 inch every hour for 6 hours what was the total snowfall that was meas
ioda

Answer:

Step-by-step explanation:

4 hours × (¼ in)/hour = 1 in

6 hours × (½ in)/hour = 3 in

Total snowfall over ten hours = 1 + 3 = 4 in

8 0
3 years ago
Evaluate the following expression at x = -4<br> -5x-1
ANTONII [103]
All u have to do is 20-1=19
3 0
3 years ago
Read 2 more answers
A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained p
Alekssandra [29.7K]

Using the z-distribution and the formula for the margin of error, it is found that:

a) A sample size of 54 is needed.

b) A sample size of 752 is needed.

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

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

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is of:

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

90% confidence level, hence\alpha = 0.9, z is the value of Z that has a p-value of \frac{1+0.9}{2} = 0.95, so z = 1.645.

Item a:

The estimate is \pi = 0.213 - 0.195 = 0.018.

The sample size is <u>n for which M = 0.03</u>, hence:

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

0.03 = 1.645\sqrt{\frac{0.018(0.982)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.018(0.982)}

\sqrt{n} = \frac{1.645\sqrt{0.018(0.982)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.018(0.982)}}{0.03}\right)^2

n = 53.1

Rounding up, a sample size of 54 is needed.

Item b:

No prior estimate, hence \pi = 0.05

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

0.03 = 1.645\sqrt{\frac{0.5(0.5)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.645\sqrt{0.5(0.5)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.5(0.5)}}{0.03}\right)^2

n = 751.7

Rounding up, a sample of 752 should be taken.

A similar problem is given at brainly.com/question/25694087

5 0
2 years ago
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