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tankabanditka [31]
3 years ago
13

Find a missing numerators in each of the following problems. a. 10/15 = /60 b. /180 = 4 / 9 c. 7/11 = /121 d. /144 = 2/6

Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
3 0

Answer:

  a) 40

  b) 80

  c) 77

  d) 48

Step-by-step explanation:

To find the missing numerator, multiply the equation by the denominator of the fraction with the missing numerator.

a) (10/15)(60) = (x/60)(60)

  40 = x

__

b) (x/180)(180) = (4/9)(180)

  x = 80

__

c) (7/11)(121) = (x/121)(121)

  77 = x

__

d) (x/144)(144) = (2/6)(144)

  x = 48

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Answer:

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Step-by-step explanation:

In 3 hours and 30 minutes Greg has earned 64.75 dollars if you take that amount(the amount earned in 3 and a half hours) and divide it by the 3 1/2 hours(the time it took to earn that amount) you get 18.50 which is the amount you earn per half hour since that graph is increasing at 30min. per 18.50 or the amount.

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If x = -2, then x2-7x+10 equals<br> 00<br> O 20<br> 028
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Three people invest at a ratio of 1:4:7. If the partnership is worth $636,000, what did each person invest?
olga_2 [115]

Answer:

See the detailed answers below

Step-by-step explanation:

Given data

Ratio=  1:4:7

combined ratio= 1+4+7= 12

Total amount= $636,000

Apply the part to all method

The first person will take

1/12= x/ 636,000

cross multiply

12x= 636,000*1

x= 636,000/12

x= 53000

The second person will take

4/12= x/ 636,000

cross multiply

12x= 636,000*4

x= 2544000/12

x= 212000

The third person will take

7/12= x/ 636,000

cross multiply

12x= 636,000*7

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x= 371000

6 0
3 years ago
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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