I will give BRAINLIEST to the correct answer Find the measure of
1 answer:
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Answer:
P = $300
r = 0.15
n = 12
$544.61 (to the nearest cent)
$524.70 (to the nearest cent)
Step-by-step explanation:
P = principal amount = $300
r = annual interest rate in decimal form = 15% = 15/100 = 0.15
n = number of times interest is compounded per unit t = 12
<u>How much she'll owe in 4 years</u>
P = 300
r = 0.15
n = 12
t = 4
= $544.61 (to the nearest cent)
<u>Yearly compounding interest rate</u>
<u>How much she'll owe in 4 years at yearly compounding interest</u>
= $524.70 (to the nearest cent)
Answer:
A) P=2.28%
B) P=28.57%
C) I expect 10 students to be unable to complete the exam in the alloted time.
Step-by-step explanation:
In order to solve this problem, we will need to find the respective z-scores. The z-scores are found by using the following formula:
Where:
z= z-score
x= the value to normalize
= standard deviation
The z-score will help us find the area below the normal distribution curve, so in order to solve this problem we need to shade the area we need to find. (See attached picture)
A) First, we find the z-score for 60 minutes, so we get:
So now we look for the z-score on our normal distribution table. Be careful with the table you are using since some tables will find areas other than the area between the mean and the desired data. The table I used finds the area between the mean and the value to normalize.
so:
A=0.4772 for a z-score of -2
since we want to find the number of students that take less than 60 minutes, we subtract that decimal number from 0.5, so we get:
0.5-0.4772=0.0228
therefore the probability that a student finishes the exam in less than 60 minutes is:
P=2.28%
B) For this part of the problem, we find the z-score again, but this time for a time of 75 minutes:
and again we look for this z-score on the table so we get:
A=0.1915 for a z-score of -0.5
Now that we got this area we subtract it from the area we found for the 60 minutes, so we get:
0.4772-0.1915=0.2857
so there is a probability of P=28.57% of chances that the students will finish the test between 60 and 75 minutes.
C) Finally we find the z-score for a time of 90 minutes, so we get:
We look for this z-score on our table and we get that:
A=0.3413
since we need to find how many students will take longer than 90 minutes to finish the test, we subtract that number we just got from 0.5 so we get:
0.5-0.3413=0.1586
this means there is a 15.86% of probabilities a student will take longer than 90 minutes. Now, since we need to find how many of the 60 students will take longer than the 90 available minutes, then we need to multiply the total amount of students by the percentage we previously found, so we get:
60*0.1586=9.516
so approximately 10 Students will be unavailable to complete the exam in the allotted time.
