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Aleksandr [31]
4 years ago
5

On three examinations, you have grades of 85, 78, and 84. There is still a final examination, which counts as one grade In order

to get an A your average must be at least 90. If you get 100 on the final, what is your numerical average? (Type an integer or a decimal)
Mathematics
1 answer:
kirill [66]4 years ago
6 0

Answer:

The average of the provided grades are 86.75

Step-by-step explanation:

Consider the provided information.

On three examinations, you have grades of 85, 78, and 84. In order to get an A your average must be at least 90.

In the last exam you get 100 marks now calculate the average by using the formula:

\frac{\text{Sum of observations}}{\text{Number of observations}}

\frac{85+78+84+100}{5}

\frac{347}{5}

86.75

86.75 is less than 90 so you will not get A.

The average of the provided grades are 86.75

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1... -14 6(3k + 5)----------------------------------------------------------------------------------------------
Lelu [443]

Answer:

-5/3, or in decimal form, -1.666.

Step-by-step explanation:

a. -146 (3k+5)     --- distribute the -146 to "3k" and "+5."

b. -438k - 730 = 0 --- get the variable k to one side of the equation.

c. -438k=730 --- divide by -438 on both sides.

d. the answer you get is 730/-438, which, in simpler terms, is -5/3, or -1.666.

<em>Hope this helps! :)</em>

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C.

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Determine which of the following angle measures are correct. Select all the apply
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number 2 and 4

Step-by-step explanation:

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An eyewitness observes a hit-and-run accident in New York City, where 95% of the cabs are yellow and 5% are blue. A police exper
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Answer:

0.1739 = 17.39% probability that the cab actually was blue

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Witness asserts the cab is blue.

Event B: The cab is blue.

Probability of a witness assessing that a cab is blue.

20% of 95%(yellow cab, witness assesses it is blue).

80% of 5%(blue cab, witness assesses it is blue). So

P(A) = 0.2*0.95 + 0.8*0.05 = 0.23

Probability of being blue and the witness assessing that it is blue.

80% of 5%. So

P(A \cap B) = 0.8*0.05 = 0.04

What is the probability that the cab actually was blue?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.04}{0.23} = 0.1739

0.1739 = 17.39% probability that the cab actually was blue

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3 years ago
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