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Dmitry_Shevchenko [17]
3 years ago
15

Rebecca has a credit line of $6,500. She had a previous balance of $398.64 and made a payment of $250. The total of her purchase

s is $1,257.99. What is Rebecca's available credit?
Mathematics
1 answer:
valentina_108 [34]3 years ago
4 0

Answer:

$5093.37

Step-by-step explanation:

To get Rebecca's available credit we need to take all of her payments and purchases and previous balance.

Previous balance = $398.64

Purchases = $1,257.99

Payment = $250

Credit Line = $6,500

Now first we add together her purchases and previous balance.

398.64 + 1,257.99

This will give us:

$1,656.63 Total deductions.

Now we subtract that to her credit line.

$6500.00 - $1656.63

$4843.37

Now we take her payment and add it to her credit line.

$4843.37 + $250

$5093.37

So Rebecca's available credit is $5093.37.

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Answer:

4.264007e+14

Step-by-step explanation:

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It has a probability of 2 out of 36 chances as there are only two ways it can happen out of 36 different combinations that can be rolled. Product means multipication also in Maths 

<span>The first probability on the first roll is that Dice 1 must be a 5 and Dice 2 must be a 6 so that 5x6=30 </span>

<span>The second probability on the second roll is that Dice 1 must be 6 while Dice 2 must be 5 so that 6x5=30 </span>

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3 years ago
18=a<br> Over 2 <br> How do you solve that as a one step
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Multiply both sides by 2
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3 years ago
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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

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