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GREYUIT [131]
2 years ago
10

You randomly choose an undergraduate student from a large state university. Let E be the event that the student is male, let F b

e the event that the student is an in-state student, and let G be the event that the student is at least 20 years old. If these three events are probabilistically independent, and if P(E) = 0.48, P(F) = 0.72, and P(G) = 0.55, which of the following is true (to the nearest three decimals)?
a. P(student is an in-state male younger than 20) = 0.156
b. P(student is an out-of-state female younger than 20) = 0.168
c. P(student is an in-state female younger than 20) = 0.156
d. P(student is an out-of-state male at least 20) = 0.060
Mathematics
1 answer:
erik [133]2 years ago
3 0

Answer:

( a ) is the right answer .

Step-by-step explanation:

P(E) = 0.48

Probability of being male = .48

Probability of being female = .52

P(F) = 0.72

Probability of being in state  = .72

Probability of being out of state  = .28

P(G) = 0.55

Probability of being in at least 20 years   = .55

Probability of being less than 20  = .45

a )

P(student is an in-state male younger than 20)

= .72 x .48 x .45 = .1555 = .156

b )

P(student is an out-of-state female younger than 20)

= .28 x .52 x .45

= .065

c )

P(student is an in-state female younger than 20)

= .72 x .52 x .45 = .168

d )

P(student is an out-of-state male at least 20)

= .28 x .48 x .55 = .074

So , ( a ) is the right answer .

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Colin invests £2350 into a savings account. The bank gives 4.2% compound interest for the first 4 years and 4.9% thereafter. How
mixer [17]
To solve this, we are going to use the compound interest formula: A=P(1+ \frac{r}{n} )^{nt}
where
A is the final amount after t years 
P is the initial investment 
r is the interest rate in decimal form 
n is the number of times the interest is compounded per year

For the first 4 years we know that: P=2350, r= \frac{4.2}{100} =0.042, t=4, and since the problem is not specifying how often the interest is communed, we are going to assume it is compounded annually; therefore, n=1. Lest replace those values in our formula:
A=P(1+ \frac{r}{n} )^{nt}
A=2350(1+ \frac{0.042}{1} )^{(1)(4)}
A=2350(1+0.042)^{4}
A=2770.38

Now, for the next 6 years the intial investment will be the final amount from our previous step, so P=2770.38. We also know that: r= \frac{4.9}{100} =0.049, t=6, and n=1. Lets replace those values in our formula one more time:
A=P(1+ \frac{r}{n} )^{nt}
A=2770.38(1+ \frac{0.049}{1})^{(1)(6)
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4 0
2 years ago
The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar
Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

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irinina [24]

Answer:

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