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Shkiper50 [21]
3 years ago
12

When Mary was born, she weighed 8 pounds. When she was 10 years old, she weighed 10 times as much. How much more did Mary weigh

when she was 10 years old than she was born?
Mathematics
2 answers:
crimeas [40]3 years ago
7 0

Answer:80 pounds


Step-by-step explanation:

You take 8×10=80

Mazyrski [523]3 years ago
6 0
She weighed 72 pounds more when she was ten years old than when she was born.


10×8=80
80-8=72
8 in the second equation equals how much she weighed when she was born, the 80 equals how much she weighed at 10.
subtract those to find how much more she weighed at 10 years old than at birth.
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90 is 75% of what number
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The time needed to complete a final examination in a particular college course is normally distributed with a mean of 77 minutes
Komok [63]

Answer:

a. The probability of completing the exam in one hour or less is 0.0783

b. The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. The number of students will be unable to complete the exam in the allotted time is 8

Step-by-step explanation:

a. According to the given we have the following:

The time for completing the final exam in a particular college is distributed normally with mean (μ) is 77 minutes and standard deviation (σ) is 12 minutes

Hence, For X = 60, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=60−77 /12

Z=−1.4167

Using the standard normal table, the probability P(Z≤−1.4167) is approximately 0.0783.

P(Z≤−1.4167)=0.0783

Therefore, The probability of completing the exam in one hour or less is 0.0783.

b. In this case For X = 75, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=75−77 /12

Z=−0.1667

Using the standard normal table, the probability P(Z≤−0.1667) is approximately 0.4338.

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is obtained as follows:

P(60<X<75)=P(Z≤−0.1667)−P(Z≤−1.4167)

=0.4338−0.0783

=0.3555

​

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. In order to compute  how many students you expect will be unable to complete the exam in the allotted time we have to first compute the Z−score of the critical value (X=90) as follows:

Z=  X−μ /σ

Z=90−77 /12

Z​=1.0833

UsING the standard normal table, the probability P(Z≤1.0833) is approximately 0.8599.

Therefore P(Z>1.0833)=1−P(Z≤1.0833)

=1−0.8599

=0.1401

​

Therefore, The number of students will be unable to complete the exam in the allotted time is= 60×0.1401=8.406

The number of students will be unable to complete the exam in the allotted time is 8

6 0
3 years ago
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