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3 years ago

Mario had 9 green 810 Brown 6 Orange and 9 blue m&m's what is the fractions of M&M's that are orange? Help please

1 answer:

6 0

Hey You!

Let's add up the total about of M&M's.

9 + 810 + 6 + 9 = 834

That is the denominator for the fraction. The numerator is the amount of orange M&M's, which have a quantity of 6.

= 6/834

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The admission fee at an amusement park is $2.50 for children and $5.40 for adults. On a certain day, 324 people entered the park

zheka24 [161]

Answer:

124 children
200 adults

Step-by-step explanation:

It usually works well to let a variable represent quantity of the highest-value contributor. Here, that is the quantity of adult tickets.

Let "a" represent the number of adult tickets sold. Then (324-a) is the number of admission fees for children. The total revenue is ...

5.40a +2.50(324 -a) = 1390

2.90a +810 = 1390 . . . . eliminate parentheses

2.90a = 580 . . . . . . . . . subtract 810

a = 200 . . . . . . . . . . . . . .divide by 2.90

The number of children admitted was 124; the number of adults admitted was 200.

8 0

3 years ago

Read 2 more answers

The CBGC pays college and high school students $5.75 per hour to tutor the younger boys and girls. If one of those tutors works

Levart [38]

Answer:

1. $35.075

2. 43.5 hours

3. 70%

4. Danny is correct

5. Abe's score out of the 30 points of the last quiz to get an overall checkup quiz grade of 80% should be 28.5

Step-by-step explanation:

The pay per hour for student tutors = $5.75

1. A tutor who worked 3.4 hours on Monday and 2.7 hours on Thursday has worked, 3.4 + 2.7 = 6.1 hours for the week, therefore

The week earning for the tutor = 6.1 × 5.75 = $35.075

2. The Club's tutoring budget per week = $250

The number of tutoring hours that can be provided per week is given by the relation;

$\frac{The \, Club's \ tutoring\ budget \ per \, week}{The \, pay \ per \, hour\, for\, student\, tutors} = \frac{\$250}{\$5.75/hour} = 43.47826 \approx 43.5 \, hours$

3. The score of 21 out of 30 as a percentage is given as follows;

$\frac{21}{30} \times 100 = 70\%$

4. Where the second score is 45 out of a total of 60 points, we have;

$\frac{45}{60} \times 100 = 75\%$

Therefore, the increase in Abe's score is 5%, therefore, Danny is correct

5. To raise his checkup quiz grade to 80% overall, then we should have;

$\frac{70 + 75 + Q}{300} =\frac{80}{100}$

Where:

Q = The percentage grade of the last quiz

(145 + Q)×100 = 80×300 = 24000

145 + Q = 240

Q = 240 - 145 = 95%

Therefore, he has to score, 95% of 30 on the last quiz, which is thus;

$\frac{95}{100} \times 30 = 28.5$

Abe's score out of the 30 points of the last quiz to get an overall checkup quiz grade of 80% = 28.5 points.

7 0

3 years ago

Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutt

sattari [20]

Answer:

b: See first attached photo

c: V = x²y

d and e: V = x(3 - 2x)²

f: 2 cubic feet

Step-by-step explanation:

a: Sketch several boxes and calculate the volumes.

b: See first attached photo a diagram of this situation

The diagram is a square. We are cutting out squares from the corners. We don't know the size of the square yet. The side lengths were 3, but now they are 3 - 2x (since each corner has one side of the square, there are 2 sides of the cut out square on each side of the larger square)

c: The equation for volume is: V = x²y

The length and width of the box are the x values, the height would be the y value

d and e: It wants the equation for the volume for our situation. The base of the box is (3 - 2x)(3 - 2x) or (3 - 2x)². The height of the box is x, so the volume is

V = x(3 - 2x)²

f: Take the derivative, find the critical values, then plug that into x and solve for the volume. See second attached photo for the work for finding the x value that maximizes the box, and the third attached photo for the evaluation of the maximum volume...