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belka [17]
4 years ago
15

Kent walked to the bus stop, then sat and waited until the bus arrived. He rode the bus for 25 minutes, then walked the last 3 b

locks to work.
Which graph best represents the scenario?

Mathematics
2 answers:
Levart [38]4 years ago
4 0

Answer:

The last graph.

Step-by-step explanation:

the one that goes a little bit up, then its flat( thats when they waited for the bus), and then rode the bus for 25 minutes, then goes up. 50-25=25.

It goes from 0 to 10. Then 10 to 20. Then 25 up to 50.

klasskru [66]4 years ago
3 0

Answer:

the 4th graph

Step-by-step explanation:

it illustrates correctly

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Nadusha1986 [10]
To solve this, you just simply divide to get a decimal and then convert to a percentage.

2.5 divided by 9.2 is equal to approximately 0.2717.

You then multiply that by one hundred to get a percentage. So the final answer is approximately 27.17%.
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A bike shop charges x dollars to rent a bike for half a day.it charges (x+40)dollars to rent for a full day.the table shows the
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Answer:

X=560

Step-by-step explanation:

600-40=560 im not sure

5 0
3 years ago
A raffle has a grand prize of a European cruise valued at $10000 with a second prize of a Rocky Point vacation valued at $700. I
lisov135 [29]

Answer:

- 3.027

Step-by-step explanation:

First price = 10000 ; second price = 700

Number of tickets sold = 11000

Ticket cost = $4

Probability that a ticket wins grand price = 1 / 11000

Probability that a ticket wins second price = 1 / 11000

X ____ 10000 _____ 700

P(x) ___ 1 / 11000 ___ 1/11000

Expected winning for a ticket buyer :

E(X) = Σx*p(x)

E(X) = (1/11000 * 10000) + (1/11000 * 700) - ticket cost

E(X) = 0.9090909 + 0.0636363 - 4

E(X) = - 3.0272728

E(X) = - 3.027

8 0
3 years ago
The following statement is false. Rewrite the statement to make it true.
Ksju [112]

we are given

\sum _{n=0}^55n\:

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n starts from 0 and goes to 5

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6 0
3 years ago
(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice
vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that \mu = 100, \sigma = 15

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{107.5 - 100}{15}

Z = 0.5

Z = 0.5 has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{130 - 100}{15}

Z = 2

Z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549

P(X \geq 1) = 1 - P(X = 0) = 0.0451

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

8 0
3 years ago
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