If we add $0.14 until we get $21.64, we would add it 154.572 times. To check this answer and make sure I’m right, do this equation: 0.14 x 154.572, then you would get 21.64.

The correct answer is 154.572 minutes.

5 0

3 years ago

Fir what values of p will the equation x^2=p have 1 real number solution why

Naddika [18.5K]

The only value of p for which the equation has exactly 1 real number solution is p=0.

If p<0, there are no real number solutions and if p>0, there are 2 real number solutions.

Hope this helps,

Brian

8 0

3 years ago

Suppose the horses in a large stable have a mean weight of 1467lbs, and a standard deviation of 93lbs. What is the probability t

krok68 [10]

Answer:

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 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:

$\mu = 1467, \sigma = 93, n = 49, s = \frac{93}{\sqrt{49}} = 13.2857$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable?

This is the pvalue of Z when X = 1467 + 9 = 1476 subtracted by the pvalue of Z when X = 1467 - 9 = 1458.

X = 1476

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

By the Central Limit Theorem

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

$Z = \frac{1476 - 1467}{13.2857}$

$Z = 0.68$

$Z = 0.68$ has a pvalue of 0.7517

X = 1458

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

$Z = \frac{1458 - 1467}{13.2857}$

$Z = -0.68$

$Z = -0.68$ has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

5 0

3 years ago