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tatiyna
3 years ago
13

You are studying for your final exam of the semester. up to this point,you recieved 3 exam scores of 76%,99%, and 86%. to reciev

e a grade of A in the class you must average exam score between 90% and 100% for all 4 exams including the final.
Find the widest range of scores that you can get on the final exam in order to recieve a grade of A for the class.Use interval notation
Mathematics
2 answers:
Levart [38]3 years ago
4 0
What you have to do is add up the amount of total grade scores. 76+99+86+x. You get 261+x. you have to get 99% or higher

anygoal [31]3 years ago
4 0

Answer: The required range of marks is,

[99 100]

Step-by-step explanation:

Let x be the percentage score on the final exam,

Since, 3 exam scores are 76%, 99%, and 86%,

Thus,

\text{The total average percentage scores in four exams}= \frac{\text{Total sum of four exams percentage}}{4}

=\frac{ 76+99+86+x}{4}

According to the question,

For getting grade A,

90 ≤ The total average percentage scores in four exams ≤ 100

90\leq \frac{ 76+99+86+x}{4}\leq 100

360 \leq 261 + x\leq 400

360 - 261 \leq x\leq 400 -261

99\leq x\leq 129

But x can not exceed to 100 %,

\implies 99\leq x\leq 100

Thus, the required range of marks in fourth exams for getting grade A is,

[99 100]

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How are the expressions 8-15 and 8-(-15)?how are they different?
zmey [24]

8 - 15 means subtract the positive number 15 from 8.

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6 0
3 years ago
Historically 80% of the monitors made by your company will pass your stringent quality control checks. 40 monitors have just com
madam [21]

Answer:

a) 10.75% probability that exactly 30 of them will pass the quality control checks.

b) Expected value is 32

Variance is 6.4

Step-by-step explanation:

For each monitor, there are only two possible outcomes. Either they will pass the quality checks, or they will not pass these checks. The probability of a monitor passing these checks is independent of other monitors. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The variance of the binomial distribution is:

V(X) = np(1-p)

80% of the monitors made by your company will pass your stringent quality control checks.

This means that p = 0.8

40 monitors have just come off the assembly line.

This means that n = 40

a) What is the probability that exactly 30 of them will pass the quality control checks?

This is P(X = 30). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 30) = C_{40,30}.(0.8)^{30}.(0.2)^{10} = 0.1075

10.75% probability that exactly 30 of them will pass the quality control checks.

b) What are the expected value and variance of the number of these monitors that will pass the quality control checks?

E(X) = np = 40*0.8 = 32

Expected value is 32

V(X) = np(1-p) = 40*0.8*0.2 = 6.4

Variance is 6.4

4 0
3 years ago
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den301095 [7]
A= 20 let me know if you need an explanation❤️, mark me branliest pls
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