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3 years ago

Which is longer, 1/4 of line A or 1/4 of line B? Explain.

Mathematics

1 answer:

rewona [7]3 years ago

6 0

Answer:

1/4 of Line A is SMALLER than 1/4 of line B.

Step-by-step explanation:

Here, two given lines are A and B.

Now,as we can see, From the two lines Line B is LONGER.

⇒ Length of line A < Length of Line B

Now as we know if we MULTIPLY or DIVIDE an inequality by a same positive number on both the sides, then the SIGN DO NOT CHANGE.

Here, divide both sides of the equation by 4, we get:

$\implies \frac{\textrm{Length of line A}}{\textrm{4}} < \frac{\textrm{Length of line B}}{\textrm{4}}\\\implies \frac{1}{4}({\textrm{Length of line A}}) < \frac{1}{4}({\textrm{Length of line B}})$

Hence, 1/4 of Line A is SMALLER than 1/4 of line B.

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a play court on the school playground is shaped like a square joined by a semicircle.the perimeter around the entire play court

Drupady [299]

Answer:

Step-by-step explanation:

Perimeter of square = 182.8 ft

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3 years ago

andrezito [222]

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a) 0.27% probability that the mean is less than 1995 milliliters.

b) 2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

Step-by-step explanation:

To solve this problem, it is important to understand 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 $s = \frac{\sigma}{\sqrt{n}}$