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2 years ago

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An airline has 83% of its flights depart on schedule. It has 68% of its flight depart and arrive on schedule. Find the probabili

ty that a flight that departs on schedule also arrives on schedule. Round the answer to two decimal places.

2 answers:

arsen [322]2 years ago

7 0

The probability that a flight that departs on schedule also arrives on schedule
83%×68%=56%

SVEN [57.7K]2 years ago

7 0

Answer:

Hence, the probability is:

0.82

Step-by-step explanation:

Let A denote the event that the flight departs on time.

Let B denote the event that the flight arrives on schedule.

Also, A∩B denotes the event that the flight that departs on time also arrives on schedule.

Let P denotes the probability of an event.

We are asked to find:

P(B|A)

We know that:

$P(B|A)=\dfrac{P(A\bigcap B)}{P(A)}$

Hence, we have:

P(A)=0.83 x.

P(A∩B)=0.68 x.

where x are the total number of flights.

Hence,

$P(B|A)=\dfrac{0.68x}{0.83x}\\\\P(B|A)=0.82$

Hence, the probability that a flight that departs on schedule also arrives on schedule is:

0.82

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John's commute to work is 20kmhr while Sheri's commute is 500mmin. Who has the fastest commute to work in mihrif 1.61km=1mi? A S

Andreyy89

Answer:

A) Sheri has the faster commute by 6.2 miles/hr.

Step-by-step explanation:

Given

John's commute to work $=20\ km/hr$

Sheri's commute to work $=500\ m/min$

$1.61\ km = 1\ mile$

John's commute to work in miles per hour = $\frac{20\ km}{1 hr}\times \frac{1\ mile}{1.61\ km}= 12.42\ miles/hr$

Sheri's commute to work in miles per hour = $\frac{500\ m}{1\ min}\times \frac{1\ km}{1000\ m}\times \frac{1\ mile}{1.61\ km}\times \frac{60\ min}{1\ hr}= 18.63\ miles/ hr$

We can see that Sheri has a faster commute.

Difference between the rates = $18.63\ miles/ hr-12.42\ miles/hr=6.21\ miles/hr\approx 6.2\ miles/hr$

∴ Sheri has the faster commute by 6.2 miles/hr.

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2 years ago

The tables represent two linear functions in a system.<br><br> What is the solution to this system?

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The solution to this system is (x, y) = (8, -22).

The y-values get closer together by 2 units for each 2-unit increase in x. The difference at x=2 is 6, so we expect the difference in y-values to be zero when we increase x by 6 (from 2 to 8).

You can extend each table after the same pattern.

In table 1, x-values increase by 2 and y-values decrease by 8.

In table 2, x-values increase by 2 and y-values decrease by 6.

The attachment shows the tables extended to x=10. We note that the y-values are the same (-22) for x=8 (as we predicted above). That means the solution is ...

... (x, y) = (8, -22)

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1 year ago

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A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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1 year ago

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fredd [130]

Let d = the length of the trail, miles

Note that
distance = speed * time
or
time = distance / speed.

The time, t₁, to travel the trail at 3 miles per hour is
t₁ = d/3 hours
The time, t₂, to travel back at 5 miles per hour is
t₂ = d/5 hours

Because the total time is 3 hours, therefore
t₁ + t₂ = 3
d/3 + d/5 = 3
d(1/3 + 1/5) = 3
d(8/15) = 3
Multiply each side by 15.
8d = 3*15 =45
d = 45/8 = 5 5/8 miles or 5.625 miles
Total distance = 2*d = 11.25 miles or 11 1/4 miles.

t₁ = 5.625/3 = 1.875 hours or 1 hour, 52.5 minutes
t₂ = 5.625/5 = 1.125 hours or 1 hour , 7.5 minutes

Answers ;
Time to travel at 3 miles per hour = 1.875 hours (1 hour, 52.5 minutes)
Time to return at 5 miles per hour = 1.125 hours (1 hour, 7.5 minutes)
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Step-by-step explanation:

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