Answer:Multiply the top equation by -3 , then add the questions .
Step-by-step explanation:
Answer:
The equations are not compatible
Step-by-step explanation:
y = -4x + 3
8x + 2y = 8
substitute for y:
8x + 2(-4x + 3) = 8
simplify:
8x - 8x + 6 = 8
6 ≠ 8
Answer:
6c³
Step-by-step explanation:
18c³ = 2 × 3 × 3 × c × c × c
24c³ = 2 × 2 × 2 × 3 × c × c × c
Now we show the common factors in bold:
18c³ = 2 × 3 × 3 × c × c × c
24c³ = 2 × 2 × 2 × 3 × c × c × c
The common factor are:
2, 3, c, c, c
GCF = 2 × 3 × c × c × c = 6c³
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.
We can set up 2 equations given the information.
Let the price of advance ticket be represented by A and same day ticket by S
A + S = 50
20A + 40S = 1700
Solve for A in the first equation by subtracting S on both sides.
U will get A = 50 - S
Now substitute 50 - S for A in the second equation.
20 (50 - S) + 40S = 1700
1000 - 20S + 40S = 1700
20S = 700
S = 35
Same day ticket costs $35 and advance ticket costs $15
