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faust18 [17]
4 years ago
5

A radio station advertises a contest with ten cash prizes totaling $5510. There is to be a $100 difference between each successi

ve prize.
Find the amounts of the least and greatest prizes the radio station will award.

a.
least: $100
greatest: $1000
b.
least: $10
greatest: $5500
c.
least: $101
greatest: $1001
d.
least: $50
greatest: $1500



Please select the best answer from the choices provided


a) A
b) B
c) C
d) D
Mathematics
1 answer:
My name is Ann [436]4 years ago
4 0

Answer:

option C

c. least: $101

greatest: $1001

Step-by-step explanation:

A radio station advertises a contest with ten cash prizes totaling $5510. There is to be a $100 difference between each successive prize.

Sum of 10 prizes = 5510

100 is the difference. there is a common difference d=100

So its a arithmetic sequence

the sum formula for arithmetic sequence is

S_n = \frac{n}{2}(2a_1 +(n-1)d)

Sn = 5510, n=10 n d= 100 we need to find out first term a1

5510 = \frac{10}{2}(2a_1 +(10-1)100)

5510 = 5 (2a1 + 900)

5510 = 10a1 + 4500

Subtract 4500 on both sides

1010= 10a1

divide by 10 on both sides

a1 = 101

so first term that is least term is 101

To find out greatest term we use formula

a_n = a_1 + (n-1) d

a(10) = 101 + (10-1)100

= 101 + 900= 1001

greatest is 1001

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Method 2:

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3 years ago
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Answer:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean \mu = 0.39 and standard deviation s = 0.0488

Step-by-step explanation:

Central Limit Theorem

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

p = 0.39, n = 100

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By the Central Limit Theorem:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean \mu = 0.39 and standard deviation s = 0.0488

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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