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stepladder [879]
3 years ago
7

John is renting a car.The car cost 36 dollars each day plus $0.50 per mile over 100 miles.John rents the car for 5 days and driv

es 180 miles
Write and expression for how much it will cost John to rent the car
Mathematics
1 answer:
Deffense [45]3 years ago
6 0

Answer:

36*5+80 =260

Step-by-step explanation:

hope this helps

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The hartford insurance company has hired you to poll a sample of adults about their car purchases. what is wrong with using tele
lawyer [7]

Answer:

Step-by-step explanation:

Given that the hartford insurance company has hired you to poll a sample of adults about their car purchases.

Here the insurance company is using convenience sampling to select customers so they resorted to taking the names from telephone directories.

But this may not be reliable since persons owning cars need not have telephones and similarly persons having telephones need not own cars

Owning a car and name appearing in telephone directory need not be associated at all

So taking random sample of names from telephone directories may not give the required accurate result since here sample cannot be assumed to represent the population of car owners.

6 0
3 years ago
Evan spent 15.89 on 7 pounds of birdseed how much did the bird seed cost per pound
Ugo [173]
Evan spent $2.27 on each pound of bird seed. To solve this divide the cost of the bird seeds by the pounds, which will lead you to this answer. Hope this helped! :)
8 0
3 years ago
Help!!! ASAP!!!!!!!!
Svet_ta [14]

Answer:

-18y = 1

y = -1/18

5x - 9(-1/18)= -2

5x + 1/2 = -4/2

5x = -5/2

x = - 25/2

6 0
3 years ago
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Sloan [31]

Answer:

I got -16x+44/3

Step-by-step explanation:

8 0
2 years ago
Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, th
LuckyWell [14K]

Answer:

Step-by-step explanation:

Given that Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that each of their offspring will develop the disease is approximately 0.25.

Let X be the no of children having this disease out of 4 occasions

Then X has two outcomes and each trial is independent of the other trial

Hence X is binomial with n =4, and p = 0.25

a) P(X=4) = 4C4 (0.25)^4 = 0.003906

b) P(X=1) = 4C1 (0.25)(0.75)^3\\= 0.4219

c) P(4th child/3 children did not) = 0.25 since independent.

4 0
3 years ago
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