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3 years ago

Please help!

Mathematics

2 answers:

TiliK225 [7]3 years ago

8 0

The sales tax rate is 7.75%.

Jean pays tax on only items marked with T.

The items marked with T are Dog Food & Hand soap.

Cost of Dog food is $12.97

Cost of Hand soap is $3.39

Total cost of Dog Food & Hand soap is 12.97 + 3.39=$16.36

The rate of sales tax is 7.75%

So sales tax on $16.36 = 7.75% of 16.36 = 0.0775 × 16.36 = $1.2679

So Jean pays a tax of $1.2679 which is $1.27 rounded to the nearest cents.

The sales tax paid by Jean on the items marked T is $1.27

olchik [2.2K]3 years ago

3 0

The tax is $1.253 hope this helps!

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LUCKY_DIMON [66]

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2 years ago

Read 2 more answers

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3 years ago

2 ordered pairs of 6x+9y=0

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Answer:

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6x+9y=0

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3 years ago

A ski lift is designed with a total load limit of 20,000 pounds. It claims a capacity of 100 persons. An expert in ski lifts thi

Yanka [14]

Answer:

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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.

For the sum of n values of a distribution, the mean is $\mu \times n$ and the standard deviation is $\sigma\sqrt{n}$

An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.

This means that $\mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300$

If the expert is right, what is the probability that a random sample of 100 independent persons will cause an overload

Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So

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

$Z = \frac{20000 - 20000}{300}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

1 - 0.5 = 0.5

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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2 years ago

On the following number line, two rational numbers are graphed.Represent the two numbers as fractions (or mixed numbers) in lowe

nordsb [41]

The first graphed rational number is at the middle of -2 and -1, thus the number represented is $-1 \frac{1}{2}$

The second graphed number is five out of six places from from 0 to 1, thus the number graphed is $0+ \frac{5}{6} = \frac{5}{6}$

The difference between the two numbers can be represented as
$-1 \frac{1}{2} - \frac{5}{6}$
or
$\frac{5}{6} -(-1 \frac{1}{2} )= \frac{5}{6} +1 \frac{1}{2}$

To find the difference,

$-1 \frac{1}{2} - \frac{5}{6} \\ \\ =-1 \frac{3+5}{6} \\ \\ =-1 \frac{8}{6} =-2 \frac{2}{6} \\ \\ =-2 \frac{1}{3}$

Also,

$\frac{5}{6} +1 \frac{1}{2} \\ \\ =1 \frac{5+3}{6} \\ \\ =1 \frac{8}{6} =2 \frac{2}{6} \\ \\ =2 \frac{1}{6}$

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3 years ago