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4 years ago

M= 1 1/2 going through (6,4) plug into y=mx=+b

Mathematics

1 answer:

Tema [17]4 years ago

3 0

M = 1.5 = 3/2
y = mx + b
b = y - mx = 4 - (3/2)6 = 4 - 9 = -5
So equation of the line is:
y = 3x/2 - 5

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Solve y3 = 125. a.y = 5 b.y = 25 c.y = 11.2 d. y = 15

Korolek [52]

Your answer is:

a. y=5

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3 years ago

Read 2 more answers

I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.

masya89 [10]

Answer:

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

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3 years ago

An English professor assigns letter grades on a test according to the following scheme. A: Top 15% of scores B: Scores below the

lisabon 2012 [21]

Answer:

59 to 66

Step-by-step explanation:

Mean test scores = u = 74.2

Standard Deviation = $\sigma$ = 9.6

According to the given data, following is the range of grades:

Grade A: 85% to 100%

Grade B: 55% to 85%

Grade C: 19% to 55%

Grade D: 6% to 19%

Grade F: 0% to 6%

So, the grade D will be given to the students from 6% to 19% scores. We can convert these percentages to numerical limits using the z scores. First we need to to identify the corresponding z scores of these limits.

6% to 19% in decimal form would be 0.06 to 0.19. Corresponding z score for 0.06 is -1.56 and that for 0.19 is -0.88 (From the z table)

The formula for z score is:

$z=\frac{x-u}{\sigma}$