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larisa86 [58]
3 years ago
14

1. Parker was able to pay for 44% of his college tuition with his scholarship. The remaining $10,054.52 he paid for with a stude

nt loan. What was the cost of Parker’s tuition?
Mathematics
1 answer:
Nutka1998 [239]3 years ago
7 0
Parker’s tuition was 14478.5088 all you got to do is add 44% to 10,054.52 and get the answer
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A craft weighs 6 kg when empty. A lemon weighs about 0.2 kg. For economical shipping the crate with lemons must weigh at least 4
Anna71 [15]
You would need 225 lemons in the crate. you just do 45 ÷ 0.2 to find the answer xx
4 0
3 years ago
Question 3-17 i really stink at math :( it just doesnt make sense to me. alsi im failing. plz help
Trava [24]

Answer:

4,

Step-by-step explanation:

graph is a sad face, therefore a is negative. so 1 and 3 are out. the c is the y intercept, and the y intercept is at tye positve side. so 4 is the correct one.

6 0
3 years ago
How many three letter permutations are possible from the letters in the word ACTION? ("ACT" is not the same as "CAT" so order is
Alinara [238K]

Answer:

120

Step-by-step explanation:

formula:

\frac{n!}{(n-h)!}=\frac{6!}{(6-3)!}=120

7 0
3 years ago
What is the probability that more than twelve loads occur during a 4-year period? (Round your answer to three decimal places.)
Nataly_w [17]

Answer:

Given that an article suggests

that a Poisson process can be used to represent the occurrence of

structural loads over time. Suppose the mean time between occurrences of

loads is 0.4 year. a). How many loads can be expected to occur during a 4-year period? b). What is the probability that more than 11 loads occur during a

4-year period? c). How long must a time period be so that the probability of no loads

occurring during that period is at most 0.3?Part A:The number of loads that can be expected to occur during a 4-year period is given by:Part B:The expected value of the number of loads to occur during the 4-year period is 10 loads.This means that the mean is 10.The probability of a poisson distribution is given by where: k = 0, 1, 2, . . ., 11 and λ = 10.The probability that more than 11 loads occur during a

4-year period is given by:1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]= 1 - 0.571665 = 0.428335 Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283Part C:The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:Therefore, the time period that must be so that the probability of no loads

occurring during that period is at most 0.3 is given by: 3.3 years

Step-by-step explanation:

8 0
3 years ago
Binomial theorem Expand (5 - y)3
Snowcat [4.5K]

Answer:

(5 - y) ^3 = 125 - 75y + 15y^2 - y^3

Step-by-step explanation:

Binomial expression

1

1. 1

1. 2. 1

1. 3. 3. 1 --------power of 3

( 5 - y) ^3

( 5 - y) (5 - y) (5 - y)

( a + b) ^3 = a^3 + 3a^2b + 3ab^2 + b^3

a = 5

b = -y

( 5 - y) ^2 = ( 5 - y) (5 - y)

= 5( 5 - y) - y(5 - y)

= 25 - 5y - 5y + y^2

=(25-10y+y^2)

( 25 - 10y + y^2)( 5 - y)

= 5(25 - 10y + y^2) - y( 25 - 10y + y^2)

= 125 - 50y + 5y^2 - 25y + 10y^2 - y^3

Collect the like terms

= 125 - 50y - 25y + 5y^2 + 10y^2 - y^3

= 125 - 75y + 15y^2 - y^3

7 0
3 years ago
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