T = 5, so after 5 years
p(t) = t^3 - 14t^2 + 20t + 120
Take derivative to find minimum:
p’(t) = 3t^2 - 28t + 10
Factor to solve for t:
p’(t) = (3t - 2)(t - 5)
0 = (3t - 2)(t - 5)
0 = 3t - 2
2 = 3t
2/3 = t
Plug 2/3 into original equation, this is a maximum. We want the minimum:
0 = t - 5
5 = t
Plug back into original:
5^3 - 14(5)^2 + 20(5) + 120
125 - 14(25) + 100 + 120
125 - 350 + 220
- 225 + 220
p(5) = -5
Answer:
0.010
Step-by-step explanation:
We solve the above question using z score formula
z = (x-μ)/σ, where
x is the raw score = 63 inches
μ is the population mean = 70 inches
σ is the population standard deviation = 3 inches
For x shorter than 63 inches = x < 63
Z score = x - μ/σ
= 63 - 70/3
= -2.33333
Probability value from Z-Table:
P(x<63) = 0.0098153
Approximately to the nearest thousandth = 0.010
Therefore, the probability that a randomly selected student will be shorter than 63 inches tall, to the nearest thousandth is 0.010.
Answer
It takes two steps to solve this problem. Step 1, you must solve the binomial squared. Then combine similar terms.
Part 1
<h3>Answer: 13</h3>
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Explanation:
We'll replace every copy of x with -3. Then use PEMDAS to simplify.
f(x) = -2x+7
f(-3) = -2(-3)+7
f(-3) = 6+7
f(-3) = 13
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Part 2
<h3>Answer: -11</h3>
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Explanation:
We work backwards in a sense compared to what part 1 did. Instead of finding f(x) based on x, we determine what x must be for a given f(x).
We'll replace f(x) with 29 and solve for x like so
f(x) = -2x+7
29 = -2x+7
-2x+7 = 29
-2x = 29-7
-2x = 22
x = 22/(-2)
x = -11
Note how if you replaced x with -11, we'd get,
f(x) = -2x+7
f(-11) = -2(-11)+7
f(-11) = 22+7
f(-11) = 29
which helps confirm we have the correct answer.
