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raketka [301]
2 years ago
12

Based on a study from the Chronicles of Flippin'' Awesomeness, the probability that Napoleon and Pedro make it to their first pe

riod class on time is 0.26. The probability that Napoleon and Pedro catch the bus is 0.25. However, the probability that they make it to their first period class on time, given that they catch the bus is 0.61. What is the probability that Napoleon and Pedro catch the bus and make it to their first period class on time
Mathematics
1 answer:
muminat2 years ago
8 0

Answer:

15.25% probability that Napoleon and Pedro catch the bus and make it to their first period class on time

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this problem, we have that:

Event A: Both Napoleon and Pedro catch the bus, so P(A) = 0.25.

Event B: Making to their first period class on time.

However, the probability that they make it to their first period class on time, given that they catch the bus is 0.61.

This means that P(B|A) = 0.61

What is the probability that Napoleon and Pedro catch the bus and make it to their first period class on time

P(B|A) = \frac{P(A \cap B)}{P(A)}

0.61 = \frac{P(A \cap B)}{0.25}

P(A \cap B) = 0.61*0.25

P(A \cap B) = 0.1525

15.25% probability that Napoleon and Pedro catch the bus and make it to their first period class on time

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What integer is closest to 31/7
Sonja [21]

Answer:

31/7 = 4 3/7.....3/7 is less then 1/2....so the closest integer is 4

Step-by-step explanation:

Please mark me brainlist if this helped

4 0
2 years ago
For a certain type of copper wire, it is knownthat, on the average, 1.5 flaws occur per millimeter.Assuming that the number of f
mario62 [17]

Answer:

The probability that no flaws occur in a certain portion of wire of length 5 millimeters =  1.1156 occur / millimeters

Step-by-step explanation:

<u>Step 1</u>:-

Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.

by  Poisson random variable given that λ = 1.5 flaws/millimeter

Poisson distribution P(X= r) = \frac{e^{-\alpha } \alpha ^{r} }{r!}

<u>Step 2:</u>-

The probability that no flaws occur in a certain portion of wire

P(X= 0) = \frac{e^{-1.5 } \(1.5) ^{0} }{0!}

On simplification we get

P(x=0) = 0.223 flaws occur / millimeters

<u>Conclusion</u>:-

The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters

5 0
2 years ago
T.J. invests $8,100 in an account
Illusion [34]

Answer:

$13,177.97

Step-by-step explanation:

Compound Interest Formula

A=P(1+\frac{r}{n} )^{nt}

A = final amount

P = initial investment

r = interest rate

n = number of times the rate is applied during a time period

t = number of times the time period has elapsed

A=8100(1+\frac{.072}{1} )^{(1)(7)} \\A=8100(1.072)^{7}\\ A=8100*(1.626909883)\\A=13177.97006

6 0
2 years ago
The house Trevor's family lives in has 6 66 people (including Trevor) and 3 33 bathrooms. In the past month, each person showere
yKpoI14uk [10]

Answer:

The amount Trevor's family pay per minute on shower water is $0.005.

Step-by-step explanation:

The cost of water is,

Cost = $0.20/liter

Amount of water used over the entire month:

Amount of water used = 72 liters

Compute the total water cost as follows:

Total water cost = Amount of water used × Cost

                           = 72 × 0.20

                           = $14.4/liter

The average number of minutes each person showered is:

Average number of minutes = 480 minutes.

The number of people in Trevor's house:

N = 6

Compute the total amount of time the 6 person showered as follows:

Total amount of time = Average number of minutes × N

                                   = 480 × 6

                                   = 2880 minutes

Compute the pay per minute on shower water as follows:

Pay per minute on water = Total water cost ÷ Total amount of time

                                          = $14.4 ÷ 2880

                                          = $0.005

Thus, the amount Trevor's family pay per minute on shower water is $0.005.

4 0
2 years ago
Show that the given set of functions is orthogonal with respect to the given weight on the prescribed interval. Find the norm of
AnnyKZ [126]

Answer/Explanation

The complete question is:

Show that the set function {1, cos x, cos 2x, . . .} is orthogonal with respect to given weight on the prescribed interval [- π, π]

Step-by-step explanation:

If we make the identification For ∅° (x) = 1 and  ∅n(x) = cos nx, we must show that ∫ lim(π) lim(-π) .∅°(x)dx = 0 , n ≠0, and ∫ lim(π) lim(-π) .∅°(x)dx = 0, m≠n.

Therefore, in the first case, we have

(∅(x), ∅(n)) ∫ lim(π) lim(-π) .∅°(x)dx = ∫ lim(π) lim(-π) cosn(x)dx

This will therefore be equal to :

1/n sin nx lim(π) lim(-π) = 1/n  [sin nπ - sin(-nπ)] = 0 , n ≠0 (In the first case)

and in the second case, we have,,

(∅(m) , ∅(n)) = ∫ lim(π) lim(-π) .∅°(x)dx

This will therefore be equal to:

∫ lim(π) lim(-π) cos mx cos nx dx

Therefore, 1/2 ∫ lim(π) lim(-π)( cos (m+n)x + cos( m-n)x dx (Where this equation represents the trigonometric function)

1/2 [ sin (m+n)x / m+n) ]+ [ sin (m-n)x / m-n) ]  lim(π) lim(-π) = 0, m ≠ n

Now, to go ahead to find the norms in the given set intervals, we have,

for  ∅°(x) = 1 we have:

//∅°(x)//² = ∫lim(π) lim(-π) dx = 2π

So therefore, //∅°(x)//² = √2π

For ∅°∨n(x)  = cos nx  , n > 0.

It then follows that,

//∅°(x)//² = ∫lim(π) lim(-π) cos²nxdx = 1/2 ∫lim(π) lim(-π) [1 + cos2nx]dx = π

Thus, for n > 0 , //∅°(x)// = √π

It is therefore ggod to note that,

Any orthogonal set of non zero functions {∅∨n(x)}, n = 0, 1, 2, . . . can be  normalized—that is, made into an orthonormal set by dividing each function by  its norm. It follows from the above equations that has been set.

Therefore,

{ 1/√2π , cosx/√π , cos2x/√π...} is orthonormal on the interval {-π, π}.

6 0
2 years ago
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