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2 years ago

Tammy earns $40,000 gross annual salary. In Tammy's jurisdiction there is a progressive tax system with rules as follows:

Mathematics

2 answers:

ololo11 [35]2 years ago

6 0

Answer:

$18,400

Step-by-step explanation:

We need to find <em>what amount's 25% is 5400??</em>

<em />

Let this amount be a, so we can say:

a*0.25 = 5400

a = 5400/0.25

a = 21,600

Now

40,000 - 21,600 = 18,400

Thus for the first 18,400 --- the tax is 0

For next 21,600 ---- tax is 25% which is 5400 tax

So amount $18,400 is tax free. x is equal to 18400

viktelen [127]2 years ago

6 0

Answer:

The value of x ix $18400.

Step-by-step explanation:

Any amounts earned up to $x are free from income tax.

Any amounts earned beyond $x are taxed at 25%.

Tammy earns $40,000 gross annual salary and pays $5,400 per annum in income tax.

We can calculate as following; Assuming x < 40000

$5400 =(40000-x)\times0.25$

=> $5400=10000-0.25x$

=> $0.25x=4600$

x = 18400

Hence, the value of x ix $18400.

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Suppose the time required for an auto shop to do a tune-up is normally distributed, with a mean of 102 minutes and a standard de

hammer [34]

Answer:

Step-by-step explanation:

Suppose the time required for an auto shop to do a tune-up is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = points scored by students

u = mean time

s = standard deviation

From the information given,

u = 102 minutes

s = 18 minutes

1) We want to find the probability that a tune-up will take more than 2hrs. It is expressed as

P(x > 120 minutes) = 1 - P(x ≤ 120)

For x = 120

z = (120 - 102)/18 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.8413

P(x > 120) = 1 - 0.8413 = 0.1587

2) We want to find the probability that a tune-up will take lesser than 66 minutes. It is expressed as

P(x < 66 minutes)

For x = 66

z = (66 - 102)/18 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.02275

P(x < 66 minutes) = 0.02275

4 0

2 years ago

Subtract 4 ounces from 5 pounds.

malfutka [58]

Answer is 4 lb.,12 oz. As one pound has 16 ounces. When we convert the given 5 lb. into ounces we multiply it by 16. 5 lb = 5 x 16 oz 5 lb = 80 oz When we subtract 4 oz from 80 oz We get 80 - 4 = 76 oz Which is 64 oz + 12 oz (16 x 4) oz + 12 oz Which will be 4 lb. + 12 oz So the answer will be 4 lb., 12 oz.

6 0

1 year ago

Legolas shoots 3 arrows at once from his bow. He has 177 arrows. How many times can Legolas shoot his bow before he needs more a

Nataliya [291]

Answer:

Step-by-step explanation:

3x=177

3x represents the # of arrow he can shoot at once every time

177 represents the total # of arrow he has

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1 year ago

A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th

GenaCL600 [577]

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .