Select the correct answer from each drop-down menu.

Pythagorean’s 14 in 15 in x

Flura [38]

Answer:

210x

Step-by-step explanation:

5 0

3 years ago

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. Step 1 of 2 : Supp

Andru [333]

Answer:

The estimate of the proportion of people who pass out at more than 6 Gs is 0.279.

Step-by-step explanation:

Estimate of the proportion of people who pass out at more than 6 Gs.

Number of people who passed out divided by the size of the sample.

We have that:

Sample of 502 people, 140 passed out at G forces greater than 6. So the estimate is:

$p = \frac{140}{502} = 0.279$

The estimate of the proportion of people who pass out at more than 6 Gs is 0.279.

8 0

2 years ago

A salesperson earns $450.20 per week plus 15% of her weekly sales. Which of the following describes the sales necessary for the

lapo4ka [179]

First we must construct an equation to model the problem. (In this case we will use an inequality instead) This is what I came up with:
450.20+0.15s>=600.10

This equation shows how if her base earnings ($450.20) are added to 15% of her sales, represented by s, then the total will be greater than or equal to $600.10
Next, we simply solve for s. (steps shown below)

1) 450.20+0.15s>=600.10 (simply restating the inequality)
2) 0.15s>=149.90 (here I isolated the variable)
3)0.15s/0.15>=149.90/0.15 (Finally I solve for s by dividing both sides by 0.15, this will isolate s on the left and leave the answer on the right)
4) s>=999.33... (here I found the total sales the salesperson would need to reach his/her goal of earning a minimum of $600.10; the 3's after the decimal are repeating so in the next step I will round up to the nearest hundredth (b/c this is what money is rounded to and if I round down he/she would make less than her goal. This means i must round up.))
5) s>=999.34 (simple rounding; once again I rounded up b/c rounding down would slightly bring the total earnings to less than the goal)

<u>Therefore, the salesperson would need his/her sales to be $999.34 in order for his/her total earnings for the week to be at least $600.10</u> (greater than or equal to $600.10)

<u>Hope this helped!</u>

4 0

3 years ago

Factorize 12x² + 15xy

kolezko [41]

Answer:

3x(4x+5y)

Step-by-step explanation:

and yh.........

4 0

3 years ago

Read 2 more answers

Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th

Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

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

By the Central Limit Theorem

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

$Z = \frac{990 - 975}{9.34}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

X = 960

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

$Z = \frac{960 - 975}{9.34}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0

3 years ago