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larisa [96]
3 years ago
7

Two trains travel toward each other from points which are 195 miles apart. They travel at rate of 25 and 40 miles an hour respec

tively. If they start at the same time, how soon will they meet?
Mathematics
1 answer:
Tems11 [23]3 years ago
6 0

Answer: it will take 3 hours for both trains to meet.

Step-by-step explanation:

Let t represent the time it will take for both trains to meet.

The two trains travel toward each other from points which are 195 miles apart. It means that after t hours, the total distance that both trains would have covered is 195 miles.

Distance = speed × time

Distance covered by the first train after t hours is

25 × t = 25t

Distance covered by the second train after t hours is

40 × t = 40t

Therefore,

25t + 40t = 195

65t = 195

t = 195/65

t = 3 hours

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Gerry just returned from a cross-country trip. The trip was 3,000 miles from his home, and his total time in the airplane for th
e-lub [12.9K]

The speed of the jet stream is 4.55 miles per hour

Step-by-step explanation:

Distance of one way trip= 3000 miles

The total distance of the round trip= 6000 miles

Total time = 11 hrs

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= 6000/11

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3 years ago
A rectangle has a length of 12 feet and a perimeter of 40 feet
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Step-by-step explanation:

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3 0
4 years ago
For the following discrete random variable X with probability distribution:
Black_prince [1.1K]

Answer:

(a) The probability distribution is shown in the attachment.

(b) The value of E (<em>Y</em>) is 7.85.

(c) The value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d) The value of P (Y ≤ 2) is 0.60.

(e) Verified that the value of E (Y) is 7.85.

Step-by-step explanation:

(a)

The random variable <em>Y</em> is defined as: Y=3X^{2}-2X+1

For <em>X</em> = {0, 1, 2, 3} the value of <em>Y</em> are:

X=0;\ Y=3\times(0)^{2}-2\times(0)+1 =1

X=1;\ Y=3\times(1)^{2}-2\times(1)+1 =2

X=2;\ Y=3\times(2)^{2}-2\times(2)+1 =9

X=3;\ Y=3\times(3)^{2}-2\times(3)+1 =22

The probability of <em>Y</em> for different values are as follows:

P (Y = 1) = P (X = 0) = 0.20

P (Y = 2) = P (X = 1) = 0.40

P (Y = 9) = P (X = 2) = 0.15

P (Y = 22) = P (X = 3) = 0.25

The probability distribution of <em>Y</em> is shown below.

(b)

The expected value of a random variable using the probability distribution table is:

E(U)=\sum[u\times P(U=u)]

Compute the expected value of <em>Y</em> as follows:

E(Y)=\sum [y\times P(Y=y)]\\=(1\times0.20)+(2\times0.40)+(9\times0.15)+(22\times0.25)\\=7.85

Thus, the value of E (<em>Y</em>) is 7.85.

(c)

Compute the expected value of <em>X</em> as follows:

E(X)=\sum [x\times P(X=x)]\\=(0\times0.20)+(1\times0.40)+(2\times0.15)+(3\times0.25)\\=1.45

Compute the expected value of <em>X</em>² as follows:

E(X^{2})=\sum [x^{2}\times P(X=x)]\\=(0^{2}\times0.20)+(1^{2}\times0.40)+(2^{2}\times0.15)+(3^{2}\times0.25)\\=3.25

Thus, the value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d)

Compute the value of P (Y ≤ 2) as follows:

P (Y\leq 2)=P(Y=1)+P(Y=2)=0.20+0.40=0.60

Thus, the value of P (Y ≤ 2) is 0.60.

(e)

The value of E (Y) is 7.85.

E(Y)=E(3X^{2}-2X+1)=3E(X^{2})-2E(X)+1

Use the values of E (X) and E (X²) computed in part (c) to compute the value of E (Y).

E(Y)=3E(X^{2})-2E(X)+1\\=(3\times 3.25)-(2\times1.45)+1\\=7.85

Hence verified.

3 0
3 years ago
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