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Anna007 [38]
3 years ago
15

Maria and John love their skateboards. One day they were on

Mathematics
2 answers:
Alchen [17]3 years ago
6 0

Given Information:  

Distance between Maria and John = 30 miles

Maria's speed = 10 mph

John's speed = 5 mph

Butterfly's speed = 30 mph

Required Information:  

How far did the butterfly travel = ?

Answer:

How far did the butterfly travel = 60 miles

Step-by-step explanation:

The speed, distance and time are related as,

Distance = speed*time

The distance that Maria has to cover = 10t miles

The distance that John has to cover = 5t miles

The total distance between them is 30 miles

Mathematically,

10t + 5t = 30

15t = 30

t = 30/15

t = 2 hours

So that means it took Maria and John 2 hours to meet.

We know that a butterfly starting on Maria's skateboard flew back and forth

between the two skateboards until they met.

So that means butterfly flew for 2 hours at a speed of 30 mph.

Then the distance covered by the butterfly is

Distance = speed*time

Distance = 30*2

Distance = 60 miles

Therefore, the butterfly traveled a distance of 60 miles.

Evgen [1.6K]3 years ago
6 0

Answer:

The butterfly traveled a distance of 69 miles.

Step-by-step explanation:

Given that:

- The distance between Maria and John = 30 miles

- Maria's speed = 10 mph

- John's speed = 5 mph

- Butterfly's speed = 30 mph

We want to find how far the butterfly traveled.

First note that Speed is the ratio of distance traveled to the time taken.

It is given as

s = D/t

where s is speed, D is distance, and t is time.

From this, we can say that

D = st

That is, distance is the product of speed and time.

Maria ran at 10mph. Let the distance she ran be D1. Then

D1 = 10×t = 10t miles

John ran at 5mph. Let the distance he ran be D2. Then

D2 = 5×t = 5t miles

The total distance between them is 30 miles. That is the sum of D1 and D2 is 30

D1 + D2 = 10t + 5t = 30

15t = 30

t = 30/15

t = 2 hours

Meaning it took Maria and John 2 hours to meet.

Now, a butterfly starting on Maria's skateboard flew back and forth between the two skateboards until they met. It flew at a speed of 30mph, and at time 2 hours, we can easily obtain the distance it covered.

D = st

D = 30×2

= 60

Distance = 60 miles

Therefore, the butterfly traveled a distance of 60 miles.

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The demand equation illustrates the price of an item and how it relates to the demand of the item.

  • The slope of the demand function is -1/2
  • The equation of the demand function is: R(x) = (300 - 10x) \times (20 + 5x)
  • The price that maximizes her revenue is: Ghc 85

From the question, we have:

Plates = 300

Price = 20

The number of plates (x) decreases by 10, while the price (y) increases by 5. The table of value is:

\begin{array}{cccccc}x & {300} & {290} & {280} & {270} & {260} \ \\ y & {20} & {25} & {30} & {35} & {40} \ \end{array}

The slope (m) is calculated using:

m = \frac{y_2 - y_1}{x_2 - x_1}

So, we have:

m = \frac{25-20}{290-300}

m = \frac{5}{-10}

m = -\frac{1}{2}

The equation of the demand is as follows:

The initial number of plates (300) decreases by 10 is represented as: (300 - 10x).

Similarly, the initial price (20) increases by 5 is represented as: (20 + 5x).

So, the demand equation is:

R(x) = (300 - 10x) \times (20 + 5x)

Open the brackets to calculate the maximum revenue

R(x) =6000 + 1500x - 200x - 50x^2

R(x) =6000 + 1300x - 50x^2

Equate to 0

6000 + 1300x - 50x^2 =0

Differentiate with respect to x

1300 - 100x =0

Collect like terms

100x =1300

Divide by 100

x =13

So, the price at maximum revenue is:

Price= 20 + 5x

Price= 20 + 5 * 13

Price= 85

In conclusion:

  • The slope of the demand function is -1/2
  • The equation of the demand function is: R(x) = (300 - 10x) \times (20 + 5x)
  • The price that maximizes her revenue is: Ghc 85

Read more about demand equations at:

brainly.com/question/21586143

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a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-
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Answer:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

M = \frac{1}{2}(Lower + Upper)

Where

Lower \to Lower class interval

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So, we have:

Class 63-65:

M = \frac{1}{2}(63 + 65) = 64

Class 66 - 68:

M = \frac{1}{2}(66 + 68) = 67

When the computation is completed, the frequency distribution will be:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

\bar x = 70.2903

\sigma = 3.5795

<em>See attachment for result of 1-VarStats</em>

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