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madam [21]
1 year ago
11

Tuition at valley view community college costs $6,750 this year. ted has $550 saved. he also received a grant for $2,500 and a s

cholarship for $1,000. he will take out student loans for the rest of the cost. how much money will ted take out in student loans this year?
Mathematics
1 answer:
xeze [42]1 year ago
7 0

ted has $550 saved. he also received a grant for $2,500 and a scholarship for $1,000. he will take out student loans for the rest of the cost. then ted will take out in student loans this year is $2700

Tuition at valley view community college amount $6,750 this year.

ted has $550 saved.

he also received a grant for $2,500

and a scholarship for $1,000.

he will take out student loans for the rest of the cost.

how much money will ted take out in student loans this year?

Total amount ted have

=550+2500+1000

= 4050

but ted have to pay college amount =6750

ted take out in student loan = 6750 -4050

                                               = 2700

ted take out in student loans this year is $ 2700

learn more about of amount here

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Answer:

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P(x=k)=\frac{57^ke^{-57}}{k!}

Step-by-step explanation:

To calculate a parameter for the given day, we have to calculate what is the average arrival rate for the day.

This can be done with the data given:

1) From 10 to 12, 8 arrival/hour:  16 expected arrivals in this period

2) From 12 to 2, 8 to 12 arrival/hour (average: 10 arrivals): 20 expected arrivals in this period.

3) From 2 to 5, from 10 to 4 arrival/hour (average: 7 arrivals): 21 expected arrivals in this period.

The total expected arrivals in a day are: 16+20+21=57 arrivals/day.

Then, the probability distribution of the number of customers that enter the store on a given day is:

P(x=k)=\frac{57^ke^{-57}}{k!}

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3 years ago
One movie is 120 minutes long. Another is 100 minutes long.
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Answer:

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Answer:

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The first thing you have to do is to look for the number of seconds in one year. You have to multiply 365 days by 86,400 seconds.

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Next, you have to know the number of seconds in 10 billion years.

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