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goldfiish [28.3K]
3 years ago
12

Alice, Ben and Carol shared a sum of money they received. Alice received 1/5 of the money. The rest of the money was divided bet

ween Ben and Carol in the ratio 1:3. If Carol received $6 more than Alice, how much money did Ben receive?
Mathematics
1 answer:
Levart [38]3 years ago
8 0

Answer:

$3

Step-by-step explanation:

The money shared by Ben and Carol was split ...

... Ben : Carol = 1 : 3

The difference of these ratio numbers is 2, and the difference in amount received is $6. Thus, each ratio unit must stand for $6/2 = $3.

... Ben received 1×$3 = $3

_____

<em>The Rest of the Story</em>

Carol received 3×$3 = $9. Together, Ben and Carol received $3 + 9 = $12, which is 4/5 of the total amount. Then the 1/5 of the total amount that Alice received is

... (1/5)×$12/(4/5) = (1/4)×$12 = $3.

The $15 of money was split ...

... Alice : Ben : Carol = $3 : $3 : $9

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3 years ago
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Identify the solution set of 3 In 4 = 2 In x.<br> -{6}<br> -{-8,8}<br> -(8)
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Option C: The solution set is \{8\}

Explanation:

The expression is 3 \ln 4=2 \ln x

Now, let us find the solution set.

Switch sides, we get,

2 \ln x=3 \ln 4

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\ln x=\frac{3 \ln 4}{2}

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Hence, the above expression becomes,

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Simplifying, we get,

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x=8

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Hence, Option C is the correct answer.

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3 years ago
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
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Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

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Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

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                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

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5 0
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Answer:

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Step-by-step explanation:

Complete question:

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