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Nutka1998 [239]
2 years ago
12

Car insurance: $12

Mathematics
1 answer:
givi [52]2 years ago
4 0

Answer:

23%

Step-by-step explanation:

First, add all prices together

120.54 + 55.99 + 65.00 = 241.53

Then see what percent 241.53 is of 1,050

23%

Hope this helps! Good luck!

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Liam wants to save $6,265 to purchase a car. if $4,000 is deposited into an account at 7.5% interest, compounded monthly, how ma
timurjin [86]
To solve for time taken we use the compound formula:
A=P(1+r/100*n)^(n*t)
where:
A=future value
P=principle
r=rate
t=time
n=number of terms

6265=4000(1+7.5/(12*100))^(12t)
solving for t we get
1.56625=1.00625^12t
12t=72.0136
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Answer: 6 years
5 0
2 years ago
A marketing research firm would like to survey undergraduate and graduate college students about whether or not they take out st
neonofarm [45]

Answer:

Objective Minimize 10x1 +15x2 + 12x3+18x4+15x5+ 21x6

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

Step-by-step explanation:

Let

X1 = # of undergraduate students from the East region,

X2 = # of graduate students from the East region,

X3 = # of undergraduate students from the Central region,

X4 = # of graduate students from the Central region,

X5 = # of undergraduate students from the West region, and

X6 = # of graduate students from the West region.

Then  the cost functions are

y1= 10x1 +15x2

y2= 12x3+18x4

y3= 15x5+ 21x6

According to the given conditions

The constraints are

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500------- A

15X2 +18X4+21X6 ≥ 400---------B

21X6 ≥ 100

X6 ≥ 100/21

X6 ≥ 4.76

Taking

X6= 5

10X1 ≤ 500

X1 ≤ 500/10

X1≤ 5

18X4 ≥ 75

X4 ≥ 75/18

X4 ≥ 4.167

Taking

X4= 5

Putting the values

15X5+ 21X6 ≥ 300

15X5+ 21(5) ≥ 300

15X5+ 105 ≥ 300

15X5 ≥ 300-105

15X5 ≥ 195

X5 ≥ 195/15

X5 ≥ 13

Putting value of X6 and X4 in B

15X2 +18X4+21X6 ≥ 400

15X2 +18(5)+21(5) ≥ 400

15X2 +195 ≥ 400

15X2  ≥ 400-195

15X2  ≥ 205

X2  ≥ 205/15

X2  ≥ 13.67

Taking X2= 14

Now putting the values in the cost equations to check whether the conditions are satisfied.

y1= 10x1 +15x2

y1= 10 (5) + 15(14)= 50 + 210= $ 260

y3= 15x5+ 21x6

y3= 15 (13) + 21(5)

y3= 195+105= $ 300

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500

5+14+x3+5+13+5≥ 1500

x3≥ 1500-42

x3≥ 1458

y2= 12x3+18x4

y2= 12 (1458)  + 18 (5)

y2= 17496 +90

y2= $ 17586

The cost can be minimized if the number of students from

                                 Undergraduate         Graduate

East Region              X1≤ 5                            X2  ≥ 13.67

Central                      X3≥ 1458                   X4 ≥ 4.167

West                          X5 ≥ 13                          X6 ≥ 4.76

This will result in the required number of students that is 1500

Constraints:

East Undergraduate must not be greater than 5

East Graduate must not be less than 13

Central Undergraduate must  be greater than 1458

Central Graduate must  be greater than 4

West Undergraduate must  be greater than 13

West Graduate must  be greater than 4

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The east region has a least cost of $260 and west region has a cost of $300.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

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(-5,15)

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Answer:

d. Yes, all of the expected counts are greater than or equal to 5

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Yes, the requirements satisfied to conduct a hypothesis test and all of the expected counts are greater than or equal to 5.

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