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3 years ago

Lyril inspects sneakers at a

Mathematics

1 answer:

NISA [10]3 years ago

4 0

Answer:

0.8

Step-by-step explanation:

Of the 60 sneakers inspected, 48 were acceptable. The experimental probability is therefore:

P = 48/60

P = 0.8

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Suppose you are given the two functions f (x) = 2x + 3 and g(x) = –x2 + 5. Composition means that you can plug g(x) into f (x).

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Mason wants to buy some pull toys for his dogs. Each toy costs $3. He knows that 1 toy costs $3, 2 toys cost $6, and 3 toys cost

Ainat [17]

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3 years ago

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3 years ago

A rectangular yard is 20 yards wide and 40 yards long. It is surrounded by a thick hedge that grows on the border of the propert

borishaifa [10]

Answer:

$W_{h}= 4.275yd$

Step-by-step explanation:

Data

wide=20yd

length=40yd

Hedge area $A_{h}$ = $171yd^{2}$

If the area of rectangle is A=base * hight and our case L*W. then:

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3 years ago

The age at which small breed dogs are fully housebroken follows a Normal distribution with mean ms = 6 months and standard devia

atroni [7]

Answer:

This measure is just 0.17 standard deviations from the mean, so we should not be surprised.

Step-by-step explanation:

Normal probability distribution

When the distribution is normal, we use 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.

If the z-score is lower than -2, or higher than 2.5, the score of X is considered unusual.

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

Let xS – xL represent the sampling distribution.

Mean s 6, means L 4. So

$\mu = 6 - 4 = 2$

Standard deviation s is 2.5, for L is 1.5. So

$\sigma = \sqrt{2.5^2+1.5^2} = 2.915$

Should we be surprised if the sample mean housebroken age for the small breed dogs is at least 2.5 months more than the sample mean housebroken age for the large breed dogs? Explain your answer.

We have to find the z-score for X = 2.5. So

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

$Z = \frac{2.5 - 2}{2.915}$

$Z = 0.17$

This measure is just 0.17 standard deviations from the mean, so we should not be surprised.

7 0

3 years ago