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monitta
3 years ago
5

The annual 2-mile fun-run is a traditional fund-raising event to support local arts and sciences activities. It is known that th

e mean and the standard deviation of finish times for this event are respectively \mu μ = 30 and \sigma σ = 5.5 minutes. Suppose the distribution of finish times is approximately bell-shaped and symmetric. Find the approximate proportion of runners who finish in under 19 minutes.
Mathematics
1 answer:
victus00 [196]3 years ago
8 0

Answer: 0.0228

Step-by-step explanation:

Given : The  mean and the standard deviation of finish times (in minutes) for this event are respectively as :-

\mu=30\\\\\sigma=5.5

If the distribution of finish times is approximately bell-shaped and symmetric, then it must be normally distributed.

Let X be the random variable that represents the finish times for this event.

z score : z=\dfrac{x-\mu}{\sigma}

z=\dfrac{19-30}{5.5}=-2

Now, the probability of runners who finish in under 19 minutes by using standard normal distribution table :-

P(X

Hence, the approximate proportion of runners who finish in under 19 minutes = 0.0228

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VashaNatasha [74]

The common factors of the numbers 18 and 30 is 6

<h3>How to determine the common factors of the numbers?</h3>

The numbers are given as

18 and 30


Express 18 and 30 as a product of its factors

So, we have

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30 = 2 * 3 * 5

Multiply the common factors in the above expression

Common factors = 2 * 3

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So, we have

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3 0
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Gnesinka [82]
What you have to do is multiply 5.12*2
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finlep [7]

Answer:

Twelve divide by six times two.

4

Step-by-step explanation:

To solve this problem, first you have to use the order of operations stands for parenthesis, exponents, multiply, divide, add, and subtract from left to right.

First, multiply & divide numbers from left to right.

\displaystyle 12\div6\times2

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Therefore, the correct answer is twelve divide by six times two or 4.

<h2>Hope this helps!</h2><h2 /><h2>Have a wonderful blessing day! :)</h2><h2 /><h2>Good luck! :)</h2>
6 0
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