Can someone please help

The table shows the price p, for renting for a d days from a car rental company?

masha68 [24]

$\bf \begin{array}{|cc|ll} \cline{1-2} \stackrel{Days}{d}&\stackrel{Price}{P}\\ \cline{1-2} 1&\$115\\ 2&\$150&\leftarrow 115+\stackrel{35(1)}{35}\\ 3&\$185&\leftarrow 115+\stackrel{35(2)}{35+35}\\ 4&\$220&\leftarrow 115+\stackrel{35(3)}{35+35+35}\\ d&&\leftarrow 115+35d\\ \cline{1-2} \end{array}~\hfill \stackrel{part~A}{P(d)=35d+115} \\\\\\ ~\hspace{34em}$

part B)

the slope is always in a linear equation, the coefficient of the independent variable, in this case namely "35".

35 or 35/1 dollars/day means

for every passing day the car is rented out, the charge is $35, so if you rent the car "d" days, you'll be charged 35d.

7 0

3 years ago

A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40

bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

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.

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 $s = \frac{\sigma}{\sqrt{n}}$ .