Please show hw you got your answer, and please show which one is the correct one. (coupon A or coupon B).

Mathematics

1 answer:

Eddi Din [679]3 years ago

7 0

Coupon A: $18 rebate on a $59 printer
Coupon B: 35% off of a $59 printer

With coupon A, the final price is 59-18 = 41 dollars
With coupon B, the final price is 59-0.35*59 = 38.35 dollars

So coupon B offers the lower price. The price difference between the coupon options is $2.65 (41-38.35 = 2.65); meaning that coupon B is the better option by $2.65

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When you subtract one negative integer from another, will your answer be greater than or less than the integer you started with?

SVEN [57.7K]

So say for example u have -7 - (-5), think of subtracting integers as adding the opposite, so ur adding the opposite of -5, the opposite of -5 is 5, so ur adding -7 and 5= -2
another one: -15 - (-18)
again, adding the opposite. -15 plus positive 18= 3

8 0

3 years ago

The mean annual income for people in a certain city is 37 thousand dollars, with a standard deviation of 28 thousand dollars. A

Aloiza [94]

Answer:

$P( 31 < \bar X< 41)$

And we can ue the z score formula given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got for the limits:

$z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515$

$z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01$

So we want to find this probability:

$P(-1.515$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the annual income of a population, and for this case we know the following info:

$\mu=37$ and $\sigma=28$ and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by: