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3 years ago

Write 3 examples of numbers that are rational.

Mathematics

1 answer:

V125BC [204]3 years ago

6 0

Answer:

4/8, -0.5, 0.75

Step-by-step explanation:

these are rational numbers because they can be expressed as the quotient or fraction of two integers

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Forensic specialists can estimate the height of a deceased person from the lengths of the person's bones. These lengths are subs

Lera25 [3.4K]

Answer:

$130.2845\leq h\leq 137.7245$

Step-by-step explanation:

Given an inequality that relates the height h, in centimeters, of an adult female and the length f, in centimeters, of her femur by the equation

$|h - (2.47f + 54.10)| \leq 3.72$

If an adult female measures her femur as 32.25 centimeters, we can determine the possible range of her height by plugging f = 32.25cm into the modelled equation as shown:

$|h - (2.47(32.25) + 54.10)| \leq 3.72\\|h - (79.9045 + 54.10)| \leq 3.72\\|h - (134.0045)| \leq 3.72\\$

If the modulus function is positive then:

$h - 134.0045 \leq 3.72\\h \leq 3.71+134.0045\\h\leq 137.7245$

If the modulus function is negative then:

$-(h - 134.0045) \leq 3.72\\-h+134.0045 \leq 3.72\\-h\leq 3.72-134.0045\\-h\leq -130.2845\\$

multiply through by -1

$-(-h)\geq -(-130.2845)\\h\geq 130.2845\\130.2845\leq h$

combining the resulting inequalities, the estimate of the possible range of heights will be $130.2845\leq h\leq 137.7245$

8 0

3 years ago

Pls help asap thank you

Sati [7]

Use the Pythagorean theorem to find the height.

Height = sqrt(26^2-10^2)

Height = 24 cm

Volume = 20^2 x 24/3

Volume = 7,200 cm^3

7 0

3 years ago

Read 2 more answers

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

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

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0

3 years ago

Compare the solutions of

MrRa [10]

Answer:

of what?

Step-by-step explanation:

4 0

3 years ago

If 10 cups of ice cream are separated into 1/5 cup sample serving how many samples can be served

cestrela7 [59]

Answer:

50 Samples

Step-by-step explanation:

Divide the total cups of ice cream (10) by the sample amount (1/5)

$\frac{10}{\frac{1}{5} }$

10 ÷ 1/5

Use the keep change flip rule.

$\frac{10}{1}*\frac{5}{1} = 50$

8 0

4 years ago

Read 2 more answers