Ask question

Ask question

All categories

2 years ago

13

The final grade in an algebra course is based on a weighted average - homework counts 10%, quizzes count 15%, tests count 50%, a

nd the final exam counts 25%. A student has a homework average of 67, a quiz average of 62, and a test average of 71. What minimum score does the student need on the final exam to get at least a 70 for the final course grade?

2 answers:

tresset_1 [31]2 years ago

8 0

Answer: 74

<u>Step-by-step explanation:</u>

Homework + Quiz + Test + Final Exam = 70

67(10%) + 62(15%) + 71(50%) + x(25%) = 70

6.7 + 9.3 + 35.5 + 0.25x = 70

51.5 + 0.25x = 70

0.25x = 18.5

x = 74

The minimum score is 74.

Drupady [299]2 years ago

8 0

<u>Answer:</u>

x = 74

<u>Step-by-step explanation:</u>

To find the minimum score a student needs to score on the final exam to earn at least a 70 for the final course grade, we will multiply the student's scores with their respective weight-age.

Supposing the minimum score in the final exam to be x, we can write an equation:

Homework + quizzes + tests + final exam = 70

67(10%) + 62(15%) + 71(50%) + x(25%) = 70

6.7 + 9.3 + 35.5 + 0.25x = 70

51.5 + 0.25x = 70

0.25x = 18.5

x = 74

You might be interested in

LAST ONE OK BUT HELP PLZ

Liula [17]

Answer:

561 Inches

Step-by-step explanation:

8.5 * 5.5 * 12 = 561

Formula for Volume:

Length x Width x Hight

4 0

3 years ago

Consider the following frequency distribution.

Zarrin [17]

Answer:

Class interval 10-19 20-29 30-39 40-49 50-59

cumulative frequency 10 24 41 48 50

cumulative relative frequency 0.2 0.48 0.82 0.96 1

Step-by-step explanation:

1.

We are given the frequency of each class interval and we have to find the respective cumulative frequency and cumulative relative frequency.

Cumulative frequency

10

10+14=24

14+17=41

41+7=48

48+2=50

sum of frequencies is 50 so the relative frequency is f/50.

Relative frequency

10/50=0.2

14/50=0.28

17/50=0.34

7/50=0.14

2/50=0.04

Cumulative relative frequency

0.2

0.2+0.28=0.48

0.48+0.34=0.82

0.82+0.14=0.96

0.96+0.04=1

The cumulative relative frequency is calculated using relative frequency.

Relative frequency is calculated by dividing the respective frequency to the sum of frequency.

The cumulative frequency is calculated by adding the frequency of respective class to the sum of frequencies of previous classes.

The cumulative relative frequency is calculated by adding the relative frequency of respective class to the sum of relative frequencies of previous classes.

4 0

3 years ago

What is (x^-2 y^10)^2/x^5 y^-3 using only positive exponents

Eddi Din [679]

Answer:

$\frac{(x^{ - 2}{y}^{10})^{2} }{ {x}^{5} {y}^{ - 3} } \\ \frac{ {x}^{ - 4} {y}^{20} }{ {x}^{5} {y}^{ - 3 } } \\ {x}^{ - 9} {y}^{23} \\ \frac{ {y}^{23} }{ {x}^{9} } \\ ans \: c$

8 0

1 year ago

Find the coordinates of X. A. (7, 4.5) B. (7, 2.5) C. (3.5, 7) D. (7, 3.5

AVprozaik [17]

Answer:

A 7,4.5

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If women's weights are

Mariulka [41]

Answer:

$P(141.8$

And we can find this probability with this difference and using the normal standard table:

$P(-0.639$

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(173.6,49.8)$

Where $\mu=173.6$ and $\sigma=49.8$

We are interested on this probability

$P(141.8$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(141.8$

And we can find this probability with this difference and using the normal standard table:

$P(-0.639$

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

8 0

2 years ago