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True [87]
2 years ago
5

The Museum of Science in Boston has an exhibit in which metal balls drop down a chute, bounce around, and wind up in one of 21 b

ins. After 1000 balls have dropped, the heights within the bins clearly follow a bell-shaped curve. The standard deviation is 3 bins.
About how many balls are in bins 1 through 17?

Mathematics
1 answer:
vodka [1.7K]2 years ago
8 0
Refer to the diagram shown below.

Assume that approximately all the data is contained under the probability distribution curve within 3 standard deviations from the mean. It is actually 99.7%.
Therefore,
bin #1 corresponds to μ-3σ,
bin #11 corresponds to μ,
bin #21 corresponds to μ + 3σ.

Let x be the random variable for bin 17.
Then by interpolation,
\frac{x-\mu}{(\mu + 3\sigma )-\mu} = \frac{17-11}{21-11} \\\\ x-\mu = 1.8\sigma

The z-score for x is
z= \frac{x-\mu}{\sigma} = \frac{1.8\sigma}{\sigma} =1.8

The probability corresponding to bin #17 is (from standard tables)
P(z≤1.8) = 0.964
Therefore if there are 1000 balls, the sum of the balls in bins 1 to 17 is
0.964 * 1000 = 96.4 ≠ 96 balls.

Answer: 96 balls

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Answer:

Option B is correct

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Step-by-step explanation:

Let x be the number.

As per the statement:

At a pizzeria, the ratio of pizzas topped with meat to pizzas topped with vegetables was 4:3 on Saturday.

The ratio of pizza topped with meat to pizzas topped with vegetables = 4 : 3

then;

the number of pizza topped with meat = 4x

and

the number pizza topped with vegetables  = 3x

It is also given that the pizzeria made a total of 84 pizzas that day.

\text{Total Pizzas} = \text{Number of pizzas topped with meat}+ \text{Number of pizzas topped with vegetables}

Substitute the given values we get;

84 = 4x+3x

Combine like terms;

84= 7x

Divide both sides by 7 we get;

12 = x

The pizza were topped with vegetables = 3x = 3(12) = 36

therefore, 36 pizzas were topped with vegetables


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2 years ago
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Answer:

The dimensions that minimize the surface are:

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Step-by-step explanation:

We have a rectangular base, that its twice as long as it is wide.

It must hold 12 yd^3 of debris.

We have to minimize the surface area, subjet to the restriction of volume (12 yd^3).

The surface is equal to:

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The volume restriction is:

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If we replace h in the surface equation, we have:

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Answer:

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Step-by-step explanation:

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