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3 years ago

11

Julie and her 7 friends roasted 34 of a bag of marshmallows on the grill after dinner. They use the rest of the bag in their hot

cocoa. What fraction of the bag does each person use in the hot cocoa?

2 answers:

andreyandreev [35.5K]3 years ago

8 0

Answer:4.8

Step-by-step explanation:

Mashcka [7]3 years ago

5 0

Answer: 4.8

Step-by-step explanation:

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Please help me I need help big time

andreyandreev [35.5K]

Answer:

MW = CD

You got it man

~~~~~

8 0

3 years ago

Answer this question

jeka57 [31]

I hope this helps you

6 0

3 years ago

A local club is arranging a charter flight to Hawaii. The cost of the trip is $586 each for 80 passengers, with a refund of $

professor190 [17]

Answer:

a) The number of passengers that will maximize the revenue received from the flight is 99.

b) The maximum revenue is $48,609.

Step-by-step explanation:

We have to analyse two cases to build a piecewise function.

If there are 80 or less passengers, we have that:

The cost of the trip is $586 for each passenger. So

$R(n) = 586n$

If there are more than 80 passengers.

There is a refund of $5 per passenger for each passenger in excess of 80. So the cost for each passenger is

$R(n) = (586 - 5(n-80))n = -5n^{2} +400n + 586n = -5n^{2} + 986n$ .

So we have the following piecewise function:

$R(n) = \left \{ {{586n}, n\leq 80 \atop {-5n^{2} + 986n}, n > 80} \right$

The maxium value of a quadratic function in the format of $y(n) = an^{2} + bn + c$ happens at:

$n_{v} = -\frac{b}{2a}$

The maximum value is:

$y(n_{v})$

So:

(a) Find the number of passengers that will maximize the revenue received from the flight.

We have to see if $n_{v}$ is higher than 80.

We have that, for $n > 80$ , $R(n) = -5n^{2} + 986n$ , so $a = -5, b = 986$

The number of passengers that will maximize the revenue received from the flight is:

$n_{v} = -\frac{b}{2a} = -\frac{986}{2(-5)} = 98.6$

Rounding up, the number of passengers that will maximize the revenue received from the flight is 99.

(b) Find the maximum revenue.

This is $R(99)$ .

$R(n) = -5n^{2} + 986n$

$R(99) = -5*(99)^{2} + 986*(99) = 48609$

The maximum revenue is $48,609.

8 0

3 years ago

C. 8tZ =p how much is 8 cups in pints?

Mnenie [13.5K]

8 cups transferred into pints is 4 pints, if that is what youre asking. /:

3 0

4 years ago

A pet store sells goldfish and hermit crabs,

Elis [28]

Answer:

$Cost = \$28$

Step-by-step explanation:

Given

Represent Goldfish with g and hermit crabs with h.

The first statement, we have:

$7g + 3h = 26$

The second statement, we have:

$4g + 5h = 28$

Required

Determine the selling price of 6 goldfish and 4 hermit crabs

The equations are:

$7g + 3h = 26$ --- (1)

$4g + 5h = 28$ --- (2)

Make g the subject in (2)

$4g + 5h = 28$

$4g = 28 - 5h$

Divide both sides by 4

$g = \frac{1}{4}(28 - 5h)$

Substitute $\frac{1}{4}(28 - 5h)$ for g in (1)

$7g + 3h = 26$

$7(\frac{1}{4}(28 - 5h)) + 3h = 26$

$\frac{7}{4}(28 - 5h) + 3h = 26$

Multiply through by 4

$4 * \frac{7}{4}(28 - 5h) + 4*3h = 26*4$

$7(28 - 5h) + 4*3h = 26*4$

Open bracket

$196 - 35h + 12h = 104$

$196 -23h = 104$

Collect Like Terms

$-23h = 104-196$

$-23h = -92$

Make h the subject

$h = \frac{-92}{-23}$

$h = \frac{92}{23}$

$h = 4$

Substitute 4 for h in $g = \frac{1}{4}(28 - 5h)$

$g = \frac{1}{4}(28 - 5*4)$

$g = \frac{1}{4}(28 - 20)$

$g = \frac{1}{4}(8)$

$g = 2$

This implies that:

1 goldfish = $2

1 hermit crab = $4

The cost of 6 goldfish and 4 hermit crabs is:

$Cost = 6g + 4h$

$Cost = 6*\$2 + 4*\$4$

$Cost = \$12 + \$16$

$Cost = \$28$

5 0

3 years ago