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melisa1 [442]
3 years ago
11

Steve's scores on 6 of his tests were 92, 78, 86, 92, 95, and 91. If he took a seventh test and raised the mean of his scores by

exactly 1 point, what was the score on the seventh test? Show your work.
Mathematics
1 answer:
monitta3 years ago
8 0

Answer:

96

Step-by-step explanation:

First we need to know the mean of the Steve's scores on 6 of his tests. Given the six scores as 92, 78, 86, 92, 95, and 91.

Mean = sum of the scores/Total test taken

Mean = 92+78+86+92+95+ 91/6

Mean = 534/6

Mean = 89

If he took the seventh test and the mean score is raised by 1 them the new mean will be expressed as;

New mean = 92+78+86+92+95+ 91+x/7 = 89+1

Where x is the new score. Note that of a new score is added, the total year taken will also change to 7

To get x;

92+78+86+92+95+ 91+x/7 = 89+1

92+78+86+92+95+ 91+x/7 = 90

534+x/7 = 90

Cross multiply

533+x = 90×7

533+x = 630

x = 630-534

x = 96

Hence the score of the seventh test is 96

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The graph is the line that goes through the points (0, -1) and (-2, 0).

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3 0
3 years ago
A test consists of 10 true/false questions. To pass the test a student must answer at least 6 questions correctly. If a student
Mademuasel [1]

Answer:

37.70% probability that the student will pass the test

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the student guesses it correctly, or he does not. The probability of a student guessing a question correctly is independent of other questions. 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.

10 true/false questions.

10 questions, so n = 10

True/false questions, 2 options, one of which is correct. So p = \frac{1}{2} = 0.5

If a student guesses on each question, what is the probability that the student will pass the test?

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

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

P(X = 6) = C_{10,6}.(0.5)^{6}.(0.5)^{4} = 0.2051

P(X = 7) = C_{10,7}.(0.5)^{7}.(0.5)^{3} = 0.1172

P(X = 8) = C_{10,8}.(0.5)^{8}.(0.5)^{2} = 0.0439

P(X = 9) = C_{10,9}.(0.5)^{9}.(0.5)^{1} = 0.0098

P(X = 10) = C_{10,10}.(0.5)^{10}.(0.5)^{0} = 0.0010

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.2051 + 0.1172 + 0.0439 + 0.0098 + 0.0010 = 0.3770

37.70% probability that the student will pass the test

8 0
3 years ago
5/5/5/5/5/5/5/55/5/5//5/5/5/5//5/5/5/5/5/5/5/5​
ElenaW [278]

Answer:

Is this a question

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally distributed with mean 1252
svp [43]
The z-score is given by the formula:
z=(x-μ)/σ
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The answer to the questions given will be as follows:
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z=(1400-1252)/129
z=1.15625
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z=(1100-1252)/129
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P(X</span>≤1100<span>)=0.1190

the answer will be:
P(1100</span>≤x≤1400<span>)=0.8770-0.1190=0.758

b]</span><span>what is the probability that a randomly selected bag contains fewer than 1000 chocolate​ chips? 
</span>z=(1000-1252)/129=-1.954
P(X≤1000)=0.0256

c] <span>what proportion of bags contains more than 1200 chocolate​ chips?
z=(1200-1252)/129
z=-0.4031
P(X</span>≥1200<span>)=1-P(X</span>≤1200<span>)=1-0.4031=0.5969

</span>
7 0
3 years ago
Solve the inequality. 4x+6&lt;3x-5 A)x&lt;1 B) x&lt;11 C) x&lt;-1 D)x&lt;-11
klasskru [66]

4x + 6 < 3x - 5 is our given inequality

4x < 3x -11 by subtracting 6 on both sides of the inequality sign

x < -11 by subtracting 3x on both sides.


Thus, x < -11 -- choice D -- is the solution.

4 0
3 years ago
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