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sashaice [31]
2 years ago
15

Marshall left home biking at a constant rate of 20 mph. One hour later, it looked like it may rain, so his friend, brett, set ou

t in his car to pick him up. Brett drove at a constant rate of 60 m[h. How many hours will marshall have been riding his bike by the time brett catches up to him
Mathematics
1 answer:
Lana71 [14]2 years ago
4 0

Answer:

0.5

Step-by-step explanation:

First, we need to find how far ahead Marshall was.  Since he had been biking at 20 mph for one hour, he had gone 20 miles.

Next, we need to find how long it will take Brett to catch up to Marshall.  In order to do this, we need to find how much faster Brett is going than Marshall.  We do this by subtracting Marshall's speed from Brett's speed.

60 - 20 = 40.  So, Brett is catching up to Marshall at 40 mph.  Now, we figure out how long it will take for someone going 40 miles per hour to go 20 miles.  We find this by dividing 40 miles per hour by 20.  This is equal to 1/2 hour.  So, it will take Brett 0.5 hours to catch up to Marshall.  This is the same as A, so A is the correct answer.

We can check our answer by seeing how far Marshall and Brett will have gone.  Marshall will have been biking for 1.5 hours, so we multiply 20 * 1.5 = 30.  Marshall went 30 miles.

Brett drove for .5 hours at 60 mph, so he went 30 miles.  Since Brett and Marshall went the same distance, our answer is correct.

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Answer:

(1) The possible outcomes are: X = {0, 1, 2, 3}.

(2) The number of times should Hartley spin a difference of 1 is 36.

(3) The number of times should Hartley spin a difference of 0 is 24.

Step-by-step explanation:

The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.

The total number of spins for each of the spinner is, <em>n</em> = 96.

(1)

The sample space of spinning both the spinners together are:

S = {(1, 1), (1, 2), (1, 3), (1, 4)

      (2, 1), (2, 2), (2, 3), (2, 4)

      (3, 1), (3, 2), (3, 3), (3, 4)

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Total = 16.

The possible outcomes are:

X = {0, 1, 2, 3}.

(2)

The sample space with the difference 1 are:

S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

n (S₁) = 6

The probability of the difference 1 is:

P(\text{Diff}=1)=\frac{n(S_{1})}{N}=\frac{6}{16}=\frac{3}{8}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 1 is:

E(\text{Diff}=1)=P(\text{Diff}=1)\times n\\\\=\frac{3}{8}\times 96\\\\=36

Thus, the number of times should Hartley spin a difference of 1 is 36.

(3)

The sample space with the difference 0 are:

S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}

n (S₂) = 4

The probability of the difference 0 is:

P(\text{Diff}=0)=\frac{n(S_{2})}{N}=\frac{4}{16}=\frac{1}{4}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 0 is:

E(\text{Diff}=0)=P(\text{Diff}=0)\times n\\\\=\frac{1}{4}\times 96\\\\=24

Thus, the number of times should Hartley spin a difference of 0 is 24.

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Answer:

<u>The true statement is D</u>

Step-by-step explanation:

The rest of the question is the attached figure and the statement options.

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============================================================

According to the graph, we will check the options:

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Because the minimum is 18

C. Over the interval [–1, 4], the local minimum is 0.  (<u>Wrong</u>)

Because the minimum is at x = 4 less than zero

D. Over the interval [4, 7], the local minimum is -7.    (<u>True</u>)

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