That would be = -p^2+4p+4
Answer: y67
Step-by-step explanation:5n − 19 + n + 7 = 144 − 6n
6n − 12 = 144 − 6n
12n = 156
n = 13
m∠z = (144−6n)°
m∠z = (144−6×13)°
m∠z = y67
Answer:
6 kg
Step-by-step explanation:
Given is weight of an object directly varies with the mass of the object .
Let the constant of proportionality be k
So,
Weight = K×(mass of the object)
For an object , weight = 45 N
mass = 9 kg
So , 45 = K×9
K = 5.
Now, for the other object ,
Weight = 30 N. K = 5 (as obtained above)
So, 30 = 5×(mass of the object)
mass of the object = 6 kg.
Answer: -1
The negative value indicates a loss
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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
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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.
Answer:
Step-by-step explanation:
Let q represent the number of quarters (the higher-value coin). Then 8-q is the number of dimes, and Jill's total amount in change is ...
0.25q +0.10(8-q) = 1.25
0.15q + 0.80 = 1.25 . . . . . . . eliminate parentheses
0.15q = 0.45 . . . . . . . . . . . . . subtract 0.80
q = 3 . . . . . . . . . . . . . . . . . . . . divide by 0.15
the number of dimes is 8-q = 8-3 = 5.
Jill has 3 quarters and 5 dimes.
