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anastassius [24]
3 years ago
5

ms. myree's bill at a restaurant was $24.20 before tax the sales tax rate was 0.05 and ms.myree also included a $5 tip how much

did she spend at th restaurant in dollars
Mathematics
1 answer:
vodka [1.7K]3 years ago
8 0
To find the tax she paid take the total amount times the rate of tax
24.20x.05=1.21
Then add the tax to the amount she paid before tax was added and the tip
$1.21+$24.20+$5=$30.41
So she spent $30.41 at the restaurant 
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A stack of boards is 24 inches high.The thickness of each board is 3/8 inch. How many boards are in the stack?
guapka [62]

Answer:

64 boards:)

Step-by-step explanation:

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2 years ago
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Es There are 30 students on the debate team and 20 students on the math team. Ten students are on both the math team and
irakobra [83]
There will be 40 students!!!
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3 years ago
A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat
zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand 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}}.

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

In this question, we have that:

\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

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

X = 205

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

Z = \frac{205 - 200}{5}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 195

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

Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

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

Z = \frac{210 - 200}{5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 195

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

Z = \frac{190 - 200}{5}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

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65% of the total expenses is for marketing, in which 500,000 is the total.

First, change the percentage into decimal

65% = 65/100 = 0.65

Next, to find the amount budgeted for marketing, multiply the total with the decimal

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$325000 is used for marketing

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6 0
3 years ago
I need help on this question ASAP.
BigorU [14]

Answer:

C

Step-by-step explanation:

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3 years ago
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