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Novay_Z [31]
3 years ago
14

X + 2y = 53x + 5y = 14Solve the system of equations.(3.1)(7.-1)(2.3/2)​

Mathematics
1 answer:
son4ous [18]3 years ago
5 0

Answer:

(3,1) because x is 3 and y is 1.

Step-by-step explanation:

x+2y=5

3x+5y=14

Multiply first equation by 3:

3x+6y=15

3x+5y=14

Subtract 3x:

1y=1

Therefore, y = 1

Plug it back into 1st equation:

x+2(1)=5

x+2=5

Therefore, x = 3

Judging from your answers, it looks like (3,1) because x is 3 and y is 1.

I hope this helped!

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[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6
viktelen [127]

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

              (b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

                  Height (in inches) of a 25 year old man is a normal random variable with mean g=71 and variance o^{2} =6.25.

To find:  (a) What percentage of 25 year old men are 6 feet, 2 inches tall

               (b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X>74in]

                           P[X>74] = P[\frac{X-g}{o} > \frac{74-71}{2.5}]

                                         = P[Z > 1.2]

                                         = 1 - P[Z ≤ 1.2]

                                         = 1 - Ф (1.2)

                                         = 1 - 0.8849

                                         = 0.1151

Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, 6ft 5in - X, 6ft]

     P[X > 6ft 5in I X > 6ft] = P[X > 77 I X > 72]

                                          = \frac{P[X > 77]}{P[ X > 72]}

                                          = \frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}]  }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }

                                          = \frac{P[Z >2.4]}{P[Z>0.4]}

                                          =  \frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }

                                          = \frac{1-0.9918}{1-0.6554}

                                          = \frac{0.0082}{0.3446}

                                          = 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

4 0
3 years ago
Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen
k0ka [10]

Let a,b,c be the randomly selected lengths. Without loss of generality, suppose a[tex]P(A + B \ge C) = P(A + B - C \ge 0)

where A,B,C are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)

so that the joint density function is

P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}

where [0,1]^3=[0,1]\times[0,1]\times[0,1] is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

a + b - c = 0

with (a,b,c)\in\Bbb R^3. This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region a+b-c\ge0, since 0+0-1=-1. So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}

5 0
2 years ago
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