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stepan [7]
3 years ago
8

A student receives test scores of 62, 83, and 91. The student's final exam score is 88 and homework score is 76. Each test is wo

rth 20% of the final grade, the final exam is 25% of the final grade, and the homework grade is 15% of the final grade. What is the student's mean score in the class?
Mathematics
1 answer:
Alborosie3 years ago
7 0

Answer:

The student's mean score in the class is 80.6

Step-by-step explanation:

we know that

To find out the student's mean score in the class, multiply each score by its worth and adds the numbers

(62+83+91)0.20+(88)0.25+(76).15=80.6

therefore

The student's mean score in the class is 80.6

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The common factor to 18xy and 24 y
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1. What is the area of the figure below? (1 point)
maw [93]

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Step-by-step explanation:

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4 0
3 years ago
Suppose a certain computer virus can enter a system through an email or through a webpage. There is a 40% chance of receiving th
DedPeter [7]

Answer:

P = 0.42

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, that is the probability of receiving this virus through the email.

-Set B, that is the probability of receiving it through the webpage.

The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that A \cap B = 0.17.

By email only

The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:

A + (A \cap B) = 0.40

A + 0.17 = 0.40

A = 0.23

where A is the probability that the system receives the virus just through the email.

The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.

By webpage only

There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:

B + (A \cap B) = 0.35

B + 0.17 = 0.35

B = 0.18

where B is the probability that the system receives the virus just through the webpage.

The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.

What is the probability that the virus does not enter the system at all?

So, we have the following probabilities.

- The virus does not enter the system: P

- The virus enters the system just by email: 23% = 0.23

- The virus enters the system just by webpage: 18% = 0.18

- The virus enters the system both by email and by the webpage: 17% = 0.17.

The sum of the probabilities is 100% = 1. So:

P + 0.23 + 0.18 + 0.17 = 1

P = 1 - 0.58

P = 0.42

There is a probability of 42% that the virus does not enter the system at all.

5 0
3 years ago
Solve (x – 3)2 = 49. Select the values of x. –46 -4 10 52
sergiy2304 [10]

After using the algebraic equation, the required value of x = 55/2.

WHAT IS ALGEBRIC EQUATION?
A mathematical statement wherein two expressions have been set equal to one another is known as an algebraic equation. A variable, coefficients, and constants make up an algebraic equation in most cases. Equations, or the equal sign, simply indicate equality. Equating each quantity with another is what equations are all about. Equations act as a scale of balance. If you've ever seen a balance scale, users know that for the scale to be deemed "balanced," an equal amount of weight must be applied to each side. The scale will tip to one side if we only add weight to one side, and the two sides will no longer be equally weighted.

(x-3)2 = 49

= x-3 = 49/2

= x = (49/2) + 3

= x = 55/2

So, the required value of x = 55/2.

To know more about algebraic equation click on the below given link

brainly.com/question/24875240

#SPJ1

4 0
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