Help me please asap ‘,:’

We need to figure out how much string would be left, after taking away the first two pieces.
We know that the first piece is 20 inches long, so we can say that there is 52-20 inches left, or 32 inches.

The second piece is between 12 and 18 inches, meaning that there would be between 32-12 and 32-18 inches left for the third piece, or 20 and 14 inches. This means that the third piece would be at least 14 inches long, but no more than 20, since we don’t have more string than that (20+12+20=52, and 20+14+18=52)

So we can say that x is greater or equal to 14, but less than or equal to 20, or:

14<=x<=20 (“<=“ is written like a normal “<“ sign with a line _ right under it)

4 0

3 years ago

Mr. Benson’s gross pay is $789.18. The amount of social security deducted is 6.2% of gross pay. How much of his gross pay will b

NeX [460]

Answer: $48.93

Step-by-step explanation: 789.18 x 0.062

8 0

3 years ago

The mean life of a television set is 119 months with a standard deviation of 14 months. If a sample of 74 televisions is randoml

irina [24]

Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

Step-by-step explanation:

To solve this question, we need 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 $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63$

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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