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3 years ago

Evaluate the function h(t)=11-3t for t=5-n

Mathematics

2 answers:

Llana [10]3 years ago

8 0

3n - 4

substitute t = 5 - n into f(t)

f(5 - n) = 11- 3(5 - n) = 11- 15 + 3n = 3n - 4

miv72 [106K]3 years ago

3 0

$h(t)=11-3(5-n)$ Substitute $5-n$ to $t$ in this equation

$h(t)=11-15+3n$ Use Distributive property

$h(t)=3n-4$ Combine Like Terms

<u><em>*(Answer)*=</em></u> $h(t)=3n-4$

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Read 2 more answers

The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0 (Think of these as

bija089 [108]

Answer:

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70

Standard Deviation, σ = 3

Sample size, n = 36

Let the average score of all pro golfers follow a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(score of the 36 golfers was between 70 and 71)

$\text{Standard error of sampling} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{1}{2}$

$P(70 \leq x \leq 71) = P(\displaystyle\frac{70 - 70}{\frac{3}{\sqrt{36}}} \leq z \leq \displaystyle\frac{71-70}{\frac{3}{\sqrt{36}}}) = P(0 \leq z \leq 2)\\\\= P(z \leq 2) - P(z \leq 0)\\= 0.977 - 0.500 = 0.477= 47.7\%$

$P(70 \leq x \leq 71) = 47.7\%$

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

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3 years ago