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marissa [1.9K]
2 years ago
13

Andrew goes to a store an buys an item that costs xx dollars. He has a coupon for 35% off, and then a 8% tax is added to the dis

counted price. Write an expression in terms of xx that represents the total amount that Andrew paid at the register
Mathematics
1 answer:
ANTONII [103]2 years ago
7 0

Answer:

xx(0.65) + xx(0.08)

Step-by-step explanation:

First step is to find the discounted price. Finding 35% less than the original xx is the same as finding 65% of it, so we multiply by 0.65. Once this is completed we need to find 8% more than that. So we add 8% of xx, which is the same as multiplying by 0.08. Hope this helps :)

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If we add, 7xy + 5yz – 3zx, 4yz + 9zx – 4y and –3xz + 5x – 2xy, then the answer is:
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Answer:

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Step-by-step explanation:

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Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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 normal distributions can be 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.

(a) If the true mean is 0.9370 with a standard deviation of 0.0090 within what interval will 99% of the sample means fail?

Samples of 34 means that n = 34

We have that \mu = 0.937, \sigma = 0.009

By the Central Limit Theorem, s = \frac{0.009}{\sqrt{34}} = 0.0015

Within what interval will 99% of the sample means fail?

Between the (100-99)/2 = 0.5th percentile and the (100+99)/2 = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

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

By the Central Limit Theorem

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

-2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = -2.575*0.0015

X = 0.933

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

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

2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = 2.575*0.0015

X = 0.941

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(b) If the true mean 0.9370 with a standard deviation of 0.0090, what is the sampling distribution of ¯X?

By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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