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3 years ago

A rectangle has a width of 9 units and a length of 40 units. what is the length of a diagonol?

1 answer:

3 0

Answers: 41 Units

You can split the rectangle into two right triangles with the side lengths 9 and 40. Then you use the Pythagorean theorem to solve for the hypotenuse (the diagonal of the rectangle). So:
C^2=9^2 + 40^2
C^2 = 1681
Square root both sides
C=41

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3 years ago

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Answer this question please; number 5... show all work thank you

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Answer:

rate of the plane in still air is 33 miles per hour and the rate of the wind is 11 miles per hour

Step-by-step explanation:

We will make a table of the trip there and back using the formula distance = rate x time

d = r x t

there

back

The distance there and back is 264 miles, so we can split that in half and put each half under d:

d = r x t

there 132

back 132

It tells us that the trip there is with the wind and the trip back is against the wind:

d = r x t

there 132 = (r + w)

back 132 = (r - w)

Finally, the trip there took 3 hours and the trip back took 6:

d = r * t

there 132 = (r + w) * 3

back 132 = (r - w) * 6

There's the table. Using the distance formula we have 2 equations that result from that info:

132 = 3(r + w) and 132 = 6(r - w)

We are looking to solve for both r and w. We have 2 equations with 2 unknowns, so we will solve the first equation for r, sub that value for r into the second equation to solve for w:

132 = 3r + 3w and

132 - 3w = 3r so

44 - w = r. Subbing that into the second equation:

132 = 6(44 - w) - 6w and

132 = 264 - 6w - 6w and

-132 = -12w so

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132 = 3r + 33 so

99 = 3r and

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3 years ago

Jessica has one piece of ribbon that is 6 2/3 yards long and one piece that is 5 1/3 yards long. What is the total length, in fe

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3 years ago

In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

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Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

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3 years ago