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Neporo4naja [7]
2 years ago
9

Marilyn has two credit cards, D and E. Card D has a balance of $691. 64, and Card E has a balance of $1,014. 22. The minimum mon

thly payment on Card D is 3. 57% of the total balance, and the minimum monthly payment on Card E is 2. 51% of the total balance. How much greater is the minimum payment on Card E than on Card D? a. $18. 85 b. $8. 10 c. $11. 24 d. $0. 77 Please select the best answer from the choices provided A B C D.
Mathematics
1 answer:
elena-14-01-66 [18.8K]2 years ago
4 0

By using basic concept of percentage we got that minimum payment on Card E is 0.77 than on Card D.

<h3>What is percentage ?</h3>

Percentage is a number or ratio expressed as a fraction of 100.

Here given that

Marilyn has two credit cards, D and E. Card D has a balance of $691. 64, and Card E has a balance of $1,014. 22. The minimum monthly payment on Card D is 3. 57% of the total balance, and the minimum monthly payment on Card E is 2. 51% of the total balance

So minimum balance of Card D is 3.57 Percentage  of $ 691.64

=3.54\times\frac{691.64}{100} =24.691548=24.69

minimum balance of Card E is  2. 51 Percentage  of  $1,014. 22.

=2.51\times\frac{1014.22}{100} =25.456922=25.46

Now we can see that minimum balance of Card E is 25.49-24.69=0.77 extra that on D

By using basic concept of percentage we got that minimum payment on Card E is 0.77 than on Card D.

To learn more about Percentage  visit : brainly.com/question/19247356

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The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de
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Answer:

1. 15.87%

2.  6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

  • Almost all the data lies within three standard deviation from the mean for a normally distributed data.
  • About 68% of data lies within one standard deviation from the mean.
  • About 95% of data lies within two standard deviations of the mean.
  • About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)

Calculation the value from standard normal z table, we have,  

P(x < 6) =0.0228 = 2.28\%

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%

84% of newborn babies weigh between 6.7 and 9.5 pounds.

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