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vichka [17]
3 years ago
12

Karyn proofreads 15 pages in 2 hours for $40.

Mathematics
2 answers:
OLEGan [10]3 years ago
8 0
Hi there.

Since she reads 15 pages in 2 hours, all we have to do to find out how much she reads in 1 hour is divide.

15/2 = 7.5

7.5 pages/hour

Same applies to the pay

40/2 = 20

$20/hour

Karyn proofreading rate is 7.5 pages per hour. Karyn receives on average $20 per hour.

~
svetlana [45]3 years ago
7 0
15/2 is her rate per hour... she proofreads 7 and a half pages in 1 hour and for 20 dollars 

20 / 7.5  is how much does she make per page or 40/15 
she makes roughly $ 2 .66
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2 years ago
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The probability that a randomly selected 2 2​-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0
Mnenie [13.5K]

Answer:

a. Probability = 0.97735

b. Probability = 0.92294

c. P(At\ Least\ One) = 1

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

P(Live) = 0.98861

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)\ and\ P(Live)

Probability = 0.98861 * 0.98861

Probability = 0.9773497321

Probability = 0.97735 <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)^n

Where n = 7

Probability = 0.98861^7

Probability = 0.92294324145

Probability = 0.92294 <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

P(At\ Least\ One) = 1 - P(None).

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

P(Not\ Live) = 0.01139

The probability that none of the 7 lives up to 3 is:

P(None) = P(Not\ Live)^7

P(None) = 0.01139^7

Substitute this value for P(None) in

P(At\ Least\ One) = 1 - P(None).

P(At\ Least\ One) = 1 - 0.01139^7

P(At\ Least\ One) = 0.99999999999997513055642436060443621

P(At\ Least\ One) = 1 ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0
2 years ago
An individual is planning a trip to a baseball game for 16 people. Of the people planning to go to the baseball game, 8 can go o
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Answer: There are 4 people who only go to the game on Saturday.

Step-by-step explanation:

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Let the number of people go on Sunday be n(B).

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According to rules, we get that

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