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mixas84 [53]
3 years ago
15

Given the following tax structure, what minimum tax would need to be assessed on shameika to make the tax progressive with respe

ct to average tax rates?
Mathematics
1 answer:
stira [4]3 years ago
6 0
It seems that you have missed some important details for us to answer this question, but hope my answer helps. So given the following tax structure, the minimum tax that would we needed to be assessed on Shameika <span> to make the tax progressive with respect to average tax rates is $3,363. Hope this helps.</span>
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I need the answers to these or for someone to tell me how to figure it out
Lorico [155]

Answer:

4. Scalene

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6. Isosceles

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9. Scalene

4 0
3 years ago
Read 2 more answers
Write -1/19, 7/4, -4/7 from least to greatest
zysi [14]

Answer:

-4/7, -1/19, 7/4

Step-by-step explanation:

The fractions converted into decimals would be -0.05, 1.75, -0.57

4 0
2 years ago
What is two and three fifths plus one and one eighth?
goblinko [34]
2 and 3/5 + 1 and 1/8
First, find the LCM of 5 and 8: 40
Next, make equivalent fractions:
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2 years ago
It has a time to failure distribution which is normal with a mean of 35,000 vehicle miles and a standard deviation of 7,000 vehi
Triss [41]

Answer:

The designed life should be of 21,840 vehicle miles.

Step-by-step explanation:

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles.

This means that \mu = 35000, \sigma = 7000

Find its designed life if a .97 reliability is desired.

The designed life should be the 100 - 97 = 3rd percentile(we want only 3% of the vehicles to fail within this time), which is X when Z has a p-value of 0.03, so X when Z = -1.88.

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

-1.88 = \frac{X - 35000}{7000}

X - 35000 = -1.88*7000

X = 21840

The designed life should be of 21,840 vehicle miles.

3 0
2 years ago
Forty-two is what percent of 70? <br><br> A.21%<br><br> B.60%<br><br> C.74%<br><br> D.167%
Ad libitum [116K]
60% is the answer you are seeking. 
5 0
3 years ago
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