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2 years ago

14

What is the approximate volume of one refraction cup? What’s its formula ?

1 answer:

Marizza181 [45]2 years ago

6 0

Answer:

Volume = 0.1696 L

The formula is Area of base of refraction cup × Depth of the refraction cup

Step-by-step explanation:

The dimensions of a refraction cup are;

Diameter = 120 mm

Depth = 30 mm

Therefore, volume of the refraction cup = Area of base × Depth

Volume of the refraction cup = (π×120²/4)/2 × 30 = 169646.003 mm³

1 L = 1000000 mm³

Therefore, 169646.003 mm³ = 169646.003/1000000 m³ = 0.1696 L

The formula is Area of base of refraction cup × Depth of the refraction cup.

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A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11. Use the TI-84 Plus calculator. (a) I

sukhopar [10]

Answer:

a) We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

b) There is a 15.97% probability that x will be between 54 and 55.

c) The 47th percentile of x is $X = 52.877$ .

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11.

This means that $\mu = 53$ .

We have to find the standard deviation of the sample, that is:

$\sigma = \frac{11}{\sqrt{45}} = 1.64$

(a) Is it appropriate to use the normal distribution to find probabilities for x?

We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

(b) Find the probability that x will be between 54 and 55.

This is the pvalue of the Z score when X = 55 subtracted by the pvalue of the Z score when X = 54.

X = 55

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55 - 53}{1.64}$

$Z = 1.22$

$Z = 1.22$ has a pvalue of 0.88877.

X = 54

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{54 - 53}{1.64}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.72907.

So, there is a 0.88877 - 0.72907 = 0.1597 = 15.97% probability that x will be between 54 and 55.

(c) Find the 47th percentile of x. Round the answer to at least two decimal places.

This is the value of X when Z has a pvalue of 0.47;

This is between $Z = -0.07$ and $Z = -0.08$ . So we use $Z = -0.075$ .

$Z = \frac{X - \mu}{\sigma}$

$-0.075 = \frac{X - 53}{1.64}$

$X = 52.877$

The 47th percentile of x is $X = 52.877$ .

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Nadya [2.5K]

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