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Kay [80]
3 years ago
7

HELP PLEASE I GIVE BRAINLIST!

Mathematics
2 answers:
MissTica3 years ago
7 0
Your answer will be B. You can draw a straight line through the graph of all the ordered pairs and F. Each input must only have one output.

Hope I helped!

Let me know if you need anything else!

~ Zoe
<span />
Jobisdone [24]3 years ago
5 0
B and F.

Hope this helps!
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Simplify the expression (k^2)^4
Andrews [41]
It would be K^{8}
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2 years ago
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A half of a class and a half of a student got A on a test. If yet another student got an A, then two-thirds of the class would h
Cerrena [4.2K]

Answer:

21

Step-by-step explanation:

6 0
2 years ago
PLEASE HELP QUICK 100 POINTS
MrRissso [65]

Answer:

Change per week: $20

Starting amount of money: $550

Step-by-step explanation:

<u>Given equation</u>:

y=550+20x

where:

  • x = number of weeks
  • y = total account balance

<u>Change per week</u>

As the variable x represents the number of weeks, its coefficient represents the change per week in the amount of money.

Therefore, the change per week in the amount of money in the account is $20.

<u>Initial Balance</u>

The starting amount of money in the account (initial balance) will be the value of y when x = 0:

\implies y=550+20(0)=550

Therefore, the starting amount of money in the account is $550.

4 0
1 year 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

4 0
2 years ago
Which triangles are similar?
lara31 [8.8K]
I’m pretty sure it would be a and b
3 0
3 years ago
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