Unfortunately, I can't do it on a graph here, but I will do it algebraically.
The solution on the graph will be the point of intersection of the two lines representing the equations.
y + 2.3 = 0.45x . . . . . (1)
-2y = 4.2x - 7.8 . . . . . (2)
From (2), y = 3.9 - 2.1x
substituting for y in (1), we have:
3.9 - 2.1x + 2.3 = 0.45x
2.55x = 6.2
x = 6.2/2.55 = 2.4
y = 3.9 - 2.1(2.4) = 3.9 - 5.04 = -1.2
Therefore, solution is (2.4, -1.2)
Answer: 615.44 cm²
Step-by-step explanation: When solving this problem the formula that you need to know is that area = π r².
The 14 cm that they are giving you is the radius of the circle, and π equals 3.14. So, in order to solve this the formula is:
Area = 3.14 x 14²
Area = 3.14 x 196
Area = 615.44 cm²
Answer:
45 x104 = 4680
Step-by-step explanation:
Answer:
-12 +26i
Step-by-step explanation:
Multiple using the FOIL Method, then combine the real and imaginary parts of the expression.
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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.
