Hey!
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Solution:
77.50 = 0.15x + 10
77.50 - 10 + 0.15x + 10 - 10
67.50 = 0.15x
0.15x = 67.50
0.15x/0.15 = 67.50/0.015
x = 450
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Check:
77.50 = 0.15(450) + 10
77.50 = 67.50 + 10
77.50 = 77.50
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Answer:
450 text messages can be sent when the monthly charge is $77.50!
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Hope This Helped! Good Luck!
First, let's work with the denominator alone:
4⁻⁴ = 1 / 4⁴ = 1/256 .
The original problem is the reciprocal of that = 256.
No one see the quadrilateral
404 error: graph not found
anyway, the graph was not included. since the sleep time was included, I will assume that the circle graph is worth 24 hours
all we need to do is to convert the percentages to fractions and multiply that by 24 to find out how many hours per activity
percent means parts out of 100 so x%=x/100
so we have
School
Eating
Sleep
Homework
Free Time
School=25%
25%=25/100=1/4
1/4 times 24=6
School: 6 hours
Eating=10%
10%=10/100=1/10
1/10 times 24=2.4
Eating: 2.4 hours
Sleep=40%
40%=40/100=4/10=2/5
2/5 times 24=48/5=9.6
Sleep: 9.6 hours
Homework=10%
10%=10/100=1/10
1/10 times 24=2.4
Homework: 2.4 hours
Free Time=15%
15%=15/100=3/30
3/20 times 24=72/20=36/10=3.6
Free Time: 3.6 Hours
Answers:
School: 6 hours
Eating: 2.4 hours
Sleep: 9.6 hours
Homework: 2.4 hours
Free Time: 3.6 Hours
Answer:
Binomial probability
Step-by-step explanation:
For each computer, there are only two possible outcomes. Either they fail, or they do not. The probability of a computer failing is independent from the probability of other computers failing. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
In this problem we have that:
To find the probability that exactly 20 of the computers will require repair on a given day, one will use what type of probability distribution
