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What is the horizontal distance from the origin to the point (6,3)

Tema [17]

Answer:

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3 0

3 years ago

Which equation represents the line that passes through points (0, 6) and (2, 0)? A.)y = negative one-third x + 2 B.)y = negative

Rudiy27

Find the slope of the line through (x1,y1) = (0,6) and (x2,y2) = (2,0)

m = (y2 - y1)/(x2 - x1)

m = (0 - 6)/(2 - 0)

m = -6/2

m = -3

The slope is -3

Since we're given the point (0,6) to be on the line, we know the y intercept is b = 6.

Plug m = -3 and b = 6 into y = mx+b to get y = -3x+6

Answer: Choice D) y = -3x+6

5 0

3 years ago

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When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

HACTEHA [7]

Answer:

(a) The value of P (X ≤ 2) is 0.8729.

(b) The value of P (X ≥ 5) is 0.0072.

(c) The value of P (1 ≤ X ≤ 4) is 0.7154.

(d) The probability that none of the 25 boards is defective is 0.2774.

(e) The expected value and standard deviation of X are 1.25 and 1.09 respectively.

Step-by-step explanation:

The random variable X is defined as the number of defective boards.

The probability that a circuit board is defective is, p = 0.05.

The sample of boards selected is of size, n = 25.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={25\choose x}0.05^{x}(1-0.05)^{25-x};\ x=0,1,2,3...$

(a)

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

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

$P(X\leq =x)=\sum\limits^{2}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.2774+0.3650+0.2305\\=0.8729$

Thus, the value of P (X ≤ 2) is 0.8729.

(b)

Compute the value of P (X ≥ 5) as follows:

P (X ≥ 5) = 1 - P (X < 5)

$=1-\sum\limits^{4}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=1-0.9928\\=0.0072$

Thus, the value of P (X ≥ 5) is 0.0072.

(c)

Compute the value of P (1 ≤ X ≤ 4) as follows:

P (1 ≤ X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\sum\limits^{4}_{x=1}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.3650+0.2305+0.0930+0.0269\\=0.7154$

Thus, the value of P (1 ≤ X ≤ 4) is 0.7154.

(d)

Compute the value of P (X = 0) as follows:

$P(X=0)={25\choose 0}0.05^{0}(1-0.05)^{25-0}=1\times 1\times 0.277389=0.2774$

Thus, the probability that none of the 25 boards is defective is 0.2774.

(e)

Compute the expected value of X as follows:

$E(X)=np=25\times 0.05=1.25$

Compute the standard deviation of X as follows:

$SD(X)=\sqrt{np(1-p)}=\sqrt{25\times 0.05\times (1-0.05)}=1.09$

Thus, the expected value and standard deviation of X are 1.25 and 1.09 respectively.

8 0

4 years ago

Here are some (simulated) data on the maximum age distribution in rabbits:

bogdanovich [222]

Let Y denote the maximum age of rabbit.

To calculate probability for Y=a from given table, we need to divide frequency corresponding to y=a with sum of frequencies.

For example P(Y=0) = $\frac{37}{37+232+429+388+225+99+58+10+6+0} =\frac{37}{1484} = 0.02493$

Like that P(y=1) = 0.1563, P(Y=2) = 0.2891, P(y=3) = 0.2615, P(Y=4) = 0.1516,

P(Y=5) = 0.0667, P(Y=6) = 0.0391,P(Y=7) = 0.0067, P(Y=8)=0.004 and P(Y=9) =0.

a) P(Y>5) = P(Y=6)+P(Y=7)+P(Y=8)+P(Y=9)

= 0.0391+0.0067+0.004+0 = 0.0498

b) P(2<Y<6) = 0.2615+0.1516+0.0667 = 0.4798

c) P(Y≥3) = 0.2615+0.1516+0.0667+0.0391+0.0067+0.004+0= 0.5296

d)P(Y<6) = 0.02493+0.1563+0.2891+0.2615+0.1516+0.0667 =0.95013

e)If we add (a) and (d), we will get 0.0498+0.95013 = 0.99993≈1

Not surprised,since this is nothing but addition of probabilities for all Y values.

That's we got 1 since numerator and denominator are same.

8 0

3 years ago

Find the sum of the first 20 terms 1.5, 1.45, 1.40, 1.35

stepan [7]

Answer:425

Step-by-step explanation:

5 0

3 years ago

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