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Elena-2011 [213]
2 years ago
11

The graph below shows the speed of a car that is driven through a town and then on a major highway. During which of the followin

g time intervals was the car stopped at a traffic light?

Mathematics
1 answer:
Varvara68 [4.7K]2 years ago
4 0

Considering the graph of the velocity of the car, it is found that the interval in which it was stopped at a traffic light was:

Between 3 and 4 minutes.

<h3>When is a car stopped at a traffic light?</h3>

When a car is stopped at a traffic light, the car is not moving, that is, it's velocity is of zero.

In this problem, the graph gives the <u>velocity as a function of time</u>, and it is at zero between 3 and 4 minutes, hence the interval in which it was stopped at a traffic light was:

Between 3 and 4 minutes.

More can be learned about the interpretation of the graph of a function at brainly.com/question/3939432

#SPJ1

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mr Goodwill [35]

Answer:

x=3 MT=36 MH=18

Step-by-step explanation:

The two triangles are the same, so 7x+8=10x-1

3x=9

x=3

MT=12x3=36

MH=36/2=18

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Answer:

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3 years ago
(2a-3)(b-a)-3a(3a+b)
ozzi

Answer: -ab - 11a^2 - 3b + 3a

Step-by-step explanation:

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3 years ago
The table below shows all outcomes for rolling 2 dice. The bold numbers 1 through 6 show the possible results for each die. All
Leokris [45]

Part I:

1. If you have to circle each outcome in the table that at least 1 die is a 4, then you have to circle row corresponding to the number 4 and column corresponding to the number 4.

2. If you have to cross out each outcome where the sum of the dice is greater than 5, then you have to cross out all not bold numbers 6, 7, 8, 9, 10, 11 and 12 in the table.

Part II:

Here you can see 36 possible outcomes.

1. The probability that at least 1 die is a 4 is:

Pr(A)=\dfrac{11}{36} - (favorable outcomes are 4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9, 4+6=10, 1+4=5, 2+4=6, 3+4=7, 5+4=9, 6+4=10).

2. The probability that the sum of the dice is greater than 5 is:

Pr(B)=\dfrac{26}{36}=\dfrac{13}{18}.

3. The probability that at least 1 die is a 4 and that the sum of the dice is greater than 5 is

Pr(A\cap B)=\dfrac{9}{36}=\dfrac{1}{4}.

4. The probability that at least 1 die is a 4 or that the sum of the dice is greater than 5 is:

Pr(A\cup B)=\dfrac{28}{36}=\dfrac{7}{9}.

4 0
3 years ago
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Alenkasestr [34]

Answer: D

Step-by-step explanation:

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LC Motors charged too much = 102

102/334 = .305..

Move the decimal 2 places to the right = 30.5

Rounds to 31%

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3 years ago
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