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Ierofanga [76]
3 years ago
14

Ann is going to rent a truck for one day. There are two companies she can choose from, and they have the following prices.

Mathematics
2 answers:
VladimirAG [237]3 years ago
5 0

Answer:

m>40

Step-by-step explanation:

So, since we know that m is for the number of miles driven we can set up the following expressions.

Company A: 93+0m

Company B: 65+0.7m

Now that we know this, we can make it an inequality.

Company A<Company B

So:

93+0m<65+0.7m

0m is 0 so our inequality is now

93+0<65+0.7m

Now, we can subtract 65 from both sides.

93-65+0<0.7m

Combine like terms.

28<0.7m

Divide both sides by 0.7

40<m

So

m>40

This means that company A charges less for mileages over 40.

ipn [44]3 years ago
4 0

Answer:

Ans: greater than 40 mileage  Company A will charge less than Company B

Step-by-step explanation:

as company A charges a fix amount $93 for any distance and company B charges variable amount starting from $65.there will be a certain mileage after that the amount charged by company B should be greater than company A.

let for greater than m mileage company A will

charge less than company B.

charge for company A for m mileage= $93

charge for company B for m mileage= $65 + $0.70*m

so,

   93  ≤ 65+0.70*m

93-65 ≤ 65+0.70*m -65

  28   ≤ 0.70*m

28*10 ≤ 7 *m

280     ≤ 7m

280/7  ≤ 7m/7

    40  ≤ m

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Answer:

No, the law of large numbers says that the proportion of yellow cards should approach the true probability after many trials.

Step-by-step explanation:

3 0
3 years ago
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by
a_sh-v [17]

Answer:

Dimension of the box is 16.1\times 7.1\times 2.45

The volume of the box is 280.05 in³.

Step-by-step explanation:          

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is L=21-2h

The width of the box is W=12-2h

The volume of the box is V=L\times W\times H

V=(21-2h)\times (12-2h)\times h

V=(21-2h)\times (12h-2h^2)

V=252h-42h^2-24h^2+4h^3

V=4h^3-66h^2+252h

To maximize the volume we find derivative of volume and put it to zero.

V'=12h^2-132h+252

0=12h^2-132h+252

Solving by quadratic formula,

h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}

h=\frac{132\pm72.99}{24}

h=2.45,8.54

Now, substitute the value of h in the volume,

V=4h^3-66h^2+252h

When, h=2.45

V=4(2.45)^3-66(2.45)^2+252(2.45)

V\approx 280.05

When, h=8.54

V=4(8.54)^3-66(8.54)^2+252(8.54)

V\approx -170.06

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is L=21-2(2.45)=16.1

The width of the box is W=12-2(2.45)=7.1

So, Dimension of the box is 16.1\times 7.1\times 2.45

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7,650÷______=7.65<br>answer______​
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Answer:

1000

3 decimals

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