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MA_775_DIABLO [31]
3 years ago
5

18. Emma rents a car from a company that rents cars by the hour. She has to pay an initial fee of $75, and then they charge her

$9 per hour. Write an equation for the total cost Cif Emma rents the car for h hours. If Emma has budgeted $250 for the rental cars, how many hours can she rent the car? Assume the car cannot be rented for part of an hour.​
Mathematics
1 answer:
love history [14]3 years ago
5 0

Step-by-step explanation:

Let h be the number of hours.

Total Cost = Fixed Costs + Variable Cost

= Initial Fee + Hourly Charge

= 75 + 9h

Given,

75 + 9h = 250 \\ 9h = 250 - 75 \\ 9h = 175 \\ h = 175 \div 9 \\  = 19 \frac{4}{9} hr

Since the car cannot be rent part of an hour, the highest possible whole number is 19 hours.

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zzz [600]

Answer:

6by6by8

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Which of the following are the zeros of the function y= 3(x-a)(x+b) in terms of a and b? ​
frosja888 [35]

The correct answer is option 2.

3 = 0

x - a = 0

x = a

x + b = 0

x = -b

We do not multiply any of these values by 3 because in a way, 3 is acting as another "zero" of the equation. Though, as shown above, it does not equal 0, it still does not apply to any of the other equations solved above.

Hope this helps!! :)

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3 years ago
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4 0
3 years ago
Factor completely 2x^3y^4-8x^2y^3+6xy^2
Maru [420]

Answer:

2 x y^2 (x y - 1) (x y - 3)

Step-by-step explanation:

2x^3y^4-8x^2y^3+6xy^2

each term contains 2xy^2 so factor that out

2xy^2(x^2y^2 -4xy+3)

then lets factor the inside

what terms multiply together to give us +3 and add to -4

-3* -1 = 3     -3+-1 =-4

2 x y^2 (x y - 1) (x y - 3)

6 0
3 years ago
The hours a week people spend answering and sending emails is normally distributed with a standard deviation of Ï = 150 minutes.
storchak [24]

Answer:

sample size n would be 149305 large

Value of n (149305) is too high, this will be the practical problem with attempting to find this confidence interval

Step-by-step explanation:

Given that;

standard deviation α = 150 min

confidence interval = 99%

since; p( -2.576 < z < 2.576) = 0.99

so z-value for 99% CI is 2.576

E = 1 minutes

Therefore

n = [(z × α) / E ]²

so we substitute

n = [(2.576 × 150) / 1 ]²

n = [ 386.4 ]²

n = 149304.96 ≈ 149305

Therefore sample size would be 149305 large

Value of n is too high, that would be the practical problem with attempting to find this confidence interval

4 0
2 years ago
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