I got 784 inches cubed, I cut the figure into two, the first one was 8×4×14=448, the second was 8×3×14=336,
therefore 448+336=784
Answer:
The coach should start recruiting players with weight 269.55 pounds or more.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 225 pounds
Standard Deviation, σ = 43 pounds
We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.
Formula:
We have to find the value of x such that the probability is 0.15
Calculation the value from standard normal z table, we have,
Thus, the coach should start recruiting players with weight 269.55 pounds or more.
Answer:
The solution is (-2, 5, 3)
Step-by-step explanation:
2x + 3y - 4z = -1
2(-2) + 3(5) - 4(3) = -1
-4 + 15 - 12 = -1
-16 + 15 = -1
-1 = -1
x - 2y + 5z = 3
-2 - 2(5) + 5(3) = 3
-2 - 10 + 15 = 3
- 12 + 15 = 3
3 = 3
-4x + y + z = 16
-4(-2) + 5 + 3 = 16
8 + 5 + 3 = 16
16 = 16
Therefore, (-2, 5, 3) is a solution.
2x + 3y - 4z = -1
2(-3.6) + 3(0.2) - 4(1.4) = -1
-7.2 + 0.6 - 5.6 = -1
-12.8 + 0.6 = -1
-12.2 = -1
x - 2y + 5z = 3
-3.6 - 2(0.2) + 5(1.4) = 3
-3.6 -0.4 + 7 = 3
-4 + 7 = 3
3 = 3
-4x + y + z = 16
-4(-3.6) + 0.2 + 1.4 = 16
14.4 + 0.2 + 1.4 = 16
16 = 16
(-3.6, 0.2, 1.4) is not a solution.
2x + 3y - 4z = -1
2(2) + 3(5) - 4(-3) = -1
4 + 15 + 12 = -1
31 = -1
x - 2y + 5z = 3
2 - 2(5) + 5(-3) = 3
2 - 10 - 15 = 3
2 - 15 = 3
13 = 3
-4x + y + z = 16
-4(2) + 5 + -3 = 16
-8 + 5 - 3 = 16
-11 + 5 = 16
-6 = 16
(2, 5, -3) is not a solution.
2x + 3y - 4z = -1
2(4) + 3(-7) - 4(-3) = -1
8 - 21 + 12 = -1
-21 + 20 = -1
-1 = -1
x - 2y + 5z = 3
4 - 2(-7) + 5(-3) = 3
4 + 14 - 15 = 3
18 - 15 = 3
3 = 3
-4x + y + z = 16
-4(4) - 7 - 3 = 16
-16 - 7 - 3 = 16
-26 = 16
(4, -7, -3) is not a solution.
Answer:
<h2>25.5 cm</h2>
Step-by-step explanation:
<h3>
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