Tax Rate Change: Suppose a certain tax rate goes from 6% to 18%. Which of the following statements is NOT true?

Mathematics

1 answer:

igor_vitrenko [27]3 years ago

4 0

Answer:

The c statement is not true.

Step-by-step explanation:

We need to check every statement to determine which one is false.

a. The old tax rate was 6% and the new one is 18%. The difference between the new and the old one will give us how much it's been increased. So 18-6=12, meaning the tax has increased 12%, so this statement is true.

b. If we multiply by 3 the old tax rate we get 6x3=18, which is the same as the new tax rate, so this statement is also true.

c. Increasing the tax rate 100% would mean multiplying by 2, since 100% is the same as the number, so adding that to the same number would double the amount. By the same logic, increasing it 200% would be the same as multiplying it by 3, and by 4 with a 300% increase. Since we already determined that the new tax was 3 times the old one, this statement is false, and the d. statement is true.

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Mekhanik [1.2K]

Answer:

A student have to score 1.28 standard deviations above the mean o be publicly recognized.

Step-by-step explanation:

If $X \sim N (\mu, \sigma^{2})$ , then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, $Z \sim N (0, 1)$ .

The distribution of these z-variates is known as the standard normal distribution.

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean.

Thus, z-scores are a type of standardized scores.

It is provided that the students who score in the top 13 percent in the National Financial Capability Challenge Exam are recognized publicly for their achievement by the Department of the Treasury.

Let x be score of the top 13%.

So, to be are recognized publicly for their achievement by the Department of the Treasury the probability of score x must be:

P (X < x) = 0.90

⇒ P (Z < z) = 0.90

The value of z for the above probability is:

z = 1.28

*Use a z-table.

Thus, a student have to score 1.28 standard deviations above the mean o be publicly recognized.