Answer:
Your answer is: 19k - 7
Simplify the expression.
Step-by-step explanation:
Hope this helped : )
Answer:
3 chickens will Mrs. Johnson put in the last shelter
Step-by-step explanation:
According to the question,
Given,Mrs. Johnson has 159 chickens that she puts in shelters at night to keep safe. She places 12 chickens in a shelter .
thus,If she put 12 chickens in each 13 shelters ( 12×13=156 chickens) adjusted in shelters .
Then, the remaining 159-156 = 3 chicken will Mrs. Johnson put in the last shelter .
Answer:
x=10, y=5
Step-by-step explanation:
For substitution, you always want to get the variable alone. The second equation (x-2y=0) will be easier to substitute into the first equation.
x-2y=0
x=2y
You can substitute 2y as x on the first equation (x+4y=30).
2y+4y=30
6y=30
y=5
Plug y=5 into any equation.
x+4(5)=30
x+20=30
x=10
I think the answer would be A not 100% sure
Answer:
0.0918
Step-by-step explanation:
We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The mean and standard deviation of average spending of sample size 25 are
μxbar=μ=95.25
σxbar=σ/√n=27.32/√25=27.32/5=5.464.
So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The z-score associated with average spending $102.5
Z=[Xbar-μxbar]/σxbar
Z=[102.5-95.25]/5.464
Z=7.25/5.464
Z=1.3269=1.33
We have to find P(Xbar>102.5).
P(Xbar>102.5)=P(Z>1.33)
P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)
P(Xbar>102.5)=0.5-0.4082
P(Xbar>102.5)=0.0918.
Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
