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horsena [70]
3 years ago
9

A truck can be rented from Company A for $70.00 a day plus $0.00 per mile. Company B charges $40.00 a day plus $0.70 per mile to

rent the same truck. Find the number of miles in a day at which the rental costs for Company A and Company B are the same?

Mathematics
1 answer:
inessss [21]3 years ago
7 0
Sent a picture of the solution to the problem (s).

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slavikrds [6]
I agree with this person that will be a good use
3 0
3 years ago
What is the name of this solid figure?
ohaa [14]

Answer:

what solid figure do u have a picture lol?

Step-by-step explanation:

4 0
3 years ago
A heated piece of metal cools according to the function c(x) = (.5)x − 11, where x is measured in hours. A device is added that
e-lub [12.9K]
16.5 degrees below the starting temp, whatever that is. If its 0, then -16.5 degrees.

The first equation would be (.5)5-11=-8.5, because the metal has been cooling for 5 hours.

The device that 'aids' in the cooling would be -5-3=-8, because it is a separate variable that cools the metal, so the amount the device cools is independent of the natural cooling amount, and the equation is independent of the natural cooling equation.
You then add -8.5 and -8, because the device has lowered 8.5 degrees and 8 degrees. This equals -16.5 degrees, or a decrease of 16.5 degrees.
7 0
3 years ago
The system of equations: 5x-4y=-3
Sophie [7]

Answer:

3)5x-4y=-3

6x+4y=14

Step-by-step explanation:

The system of equations is given

5x-4y=-3

3x+2y=7

Multiplying both sides of the lower equation by 2, we get

6x + 4y = 14

That means your answer 3)

5 0
3 years ago
A study investigated whether price affects people's judgment. Twenty people each tasted six cabernet sauvignon wines and rated h
patriot [66]

Answer:

t=5.5080( to 3 d.p)

Step-by-step explanation:

From the data given,

n =20

Deviation= 34/20= 1.7

Standard deviation (sd)= 1.3803(√Deviation)

Standard Error = sd/√n

= 1.3803/V20 = 0.3086

Test statistic is:

t = deviation /SE

= 1.7/0.3086 = 5.5080

ndf = 20 - 1 = 19

alpha = 0.01

One Tailed - Right Side Test

From Table, critical value of t =2.5395

Since the calculated value of t = 5.5080 is greater than critical value of t = 2.5395, the difference is significant. Reject null hypothesis.

t score = 5.5080

ndf = 19

One Tail - Right side Test

By Technology, p - value = 0.000

Since p - value is less than alpha , reject null hypothesis.

Conclusion:

From the result obtained it can be concluded that ,the data support the claim that the mean rating assigned to the wine when the cost is described as $90 is greater than the mean rating assigned to the wine when the cost is described as $10.

5 0
3 years ago
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