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SSSSS [86.1K]
2 years ago
9

Find P(4). 1/4 1/2 1/8

Mathematics
1 answer:
Lena [83]2 years ago
7 0

There are 8 total numbers with one 4

The probability of getting a 4 would be number of 4's / total numbers.

P(4) = 1/8

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A child got 16.67/20 on a test. what is their grade in percent form?​
solong [7]

Answer:

83.4% or 83%

Step-by-step explanation:

when trying to find a percent, divide the numerator (# of correct answers) by the denominator (# of total questions possible to get correct)

Which would be 16.67 divided by 20 which equals .8335. Move the decimal point over two placements which would be 83.35. I rounded it up to 83.4. If trying to find a whole number not a decimal it would be 84%. Depending on the grading scale, it would be a B.

3 0
3 years ago
Brody is taking a multiple choice test with a total of 20 points available. Each
Kaylis [27]
  1. Answer:

it is going to be 15fh5cu6adhfy566s

6 0
3 years ago
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You are comparing to wireless services. Verizon changes $35 a month and $5 per gigabyte of data.
ikadub [295]
A) Gigabytes = x
B) 35+5x = 25+10x
10 = 5x
2 = x
C) After 2 gigabytes, the wireless services will cost the same ($45)
7 0
3 years ago
The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo
muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: {\displaystyle P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}}

Where \lambda is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}

Substitute the respective values in the above formula.

P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})

P(k > 2)= 1 - (0.44596)

P(k > 2)=0.55404

Hence, the required probability is 0.55404.

4 0
3 years ago
What is the range of the function shown on the graph above?
nikklg [1K]

Answer:

D: 0 < y < 7

Step-by-step explanation:

The domain is the input (i.e. the x values)

The range is the output (i.e. the y values)

So for the range, look at the y-coordinates of the end points of the line: (-6, 0) and (9, 7)  ⇒  0 < y < 7

4 0
3 years ago
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