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3 years ago

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Mathematics

1 answer:

Anton [14]3 years ago

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Using partial product, are the products 48 and 64

shusha [124]

For 49 It's 4*12 and 2*24 and 1*48 and for 64 it's 8*8 and 2*32

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3 years ago

The Butlers bought a $332,000 house. They made a down payment of 42,000 and took out a mortgage for the rest. Over the course of

Zarrin [17]

Answer:

Step-by-step explanation:

(a)Mortgage = 332000 - 42000 = 290000

15 * 12 = 180 months

Total monthly payments = 180 * 2447.19 =440494.2

Total = 440494.2 + 42000 = 482494.2

(b) Interest = 482494.2 -290000

= 192494.2

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3 years ago

Read 2 more answers

Which situation relates to the graph?

enyata [817]

D because there are 10 gallons of gas remaining as you can see on the y axis and 10 gallons is being used for each 10 miles because 10 x 10 is 100 and the graph is counting up by 10

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2 years ago

Solve the following equation for h.<br> r+3Q/h=t

Arada [10]

Answer:

$\bold{h=\dfrac{3Q}{t-r}}$

Step-by-step explanation:

$\bold{r+\frac{3Q}h=t}\\-r\qquad\ -r\\\bold{\frac{3Q}h=t-r}\\\cdot h\qquad \cdot h\\ \bold{3Q=(t-r)h}\\^{\div(t-r)\quad\div(t-r)}\\\bold{\dfrac{3Q}{t-r}=h}$

Or, if you mean (r+3Q)/h=t:

$\bold{\frac{r+3Q}h=t}\\{}\ \ \cdot h\quad\ \cdot h\\\bold{r+3Q=ht}\\{}\ \ \div t\qquad \div t\\\bold{\dfrac{r+3Q}{t}=h}\\\\\bold{h=\dfrac{r+3Q}{t}}$

4 0

3 years ago

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