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Gekata [30.6K]
3 years ago
5

All exponential functions show growth over time. T or F

Mathematics
1 answer:
madam [21]3 years ago
3 0
False, ther is decay functions example

f(x)=200+(1/3)^x
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Printing a newsletter cost $1.50 per copy plus $450 in printers fee. The copies are sold for $3 each. How many copies of the new
e-lub [12.9K]
1.50x + 450 = 3x
450 = 3x - 1.50x
450 = 1.50x
450/1.50 = x
300 = x

300 copies must be sold to break even
6 0
3 years ago
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Suppose your friends parents invest $15,000 in an account paying 6% compounded annually. What will be the balance after 9 years
JulijaS [17]
A = 15000(1.04)6 .........You can put this into your calculator such that it becomes
A = $18,979.79
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2 years ago
Complete the transformation below:
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Answer:

a c

Step-by-step explanation:

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2 years ago
What is the probabibility of rolling a sum of 10 when rolling two number cubes?
Alja [10]

Answer: 1 over 12

Step-by-step explanation:

8 0
3 years ago
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You roll a pair of honest dice. If you roll a total of 7, you win $18; if you roll a total of 11, you win $54; if you roll any o
amm1812
Answer: -1
The negative value indicates a loss

============================================================

Explanation:

Define the three events
A = rolling a 7
B = rolling an 11
C = roll any other total (don't roll 7, don't roll 11)

There are 6 ways to roll a 7. They are
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
Use this to compute the probability of rolling a 7
P(A) = (number of ways to roll 7)/(number total rolls) = 6/36 = 1/6
Note: the 36 comes from 6*6 = 36 since there are 6 sides per die

There are only 2 ways to roll an 11. Those 2 ways are:
5+6 = 11
6+5 = 11
The probability for event B is P(B) = 2/36 = 1/18

Since there are 6 ways to roll a "7" and 2 ways to roll "11", there are 6+2 = 8 ways to roll either event. 
This leaves 36-8 = 28 ways to roll anything else
P(C) = 28/36 = 7/9

-----------------------------

In summary so far,
P(A) = 1/6
P(B) = 1/18
P(C) = 7/9

The winnings for each event, let's call it W(X), represents the prize amounts.
Any losses are negative values
W(A) = amount of winnings if event A happens 
W(B) = amount of winnings if event B happens
W(C) = amount of winnings if event C happens
W(A) = 18
W(B) = 54
W(C) = -9

Multiply the probability P(X) values with the corresponding W(X) values
P(A)*W(A) = (1/6)*(18) = 3
P(B)*W(B) = (1/18)*(54) = 3
P(C)*W(C) = (7/9)*(-9) = -7

Add up those results
3+3+(-7) = -1

The expected value for this game is -1.
The player is expected to lose on average 1 dollar per game played.


Note: because the expected value is not 0, this is not a fair game.


7 0
3 years ago
Read 2 more answers
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