Write an expression so that the product of a whole number and a decimal is 0.32.

Omg pls help and include work if you can. please simplify

Kruka [31]

rad 128 can be broke into 64*2

square root of 64 is 8, times -5,

which leaves -40rad2

7 0

2 years ago

Can anyone help me answer this?

netineya [11]

We need to know how much tax she must pay based on her taxable income ,
Since un the table it states that taxable incomes that range between $0-$132000have a tax rate of 18% of each $1 .
We already know her taxable income that is = 129000, which ranges between 0 and 132000.
Hence we know how much tax she must pay .
First , we need to calculate how much 18% of $1 is
= 18/100 x 1 = $0.18
Per annum/year = $0.18 x $129,000 = $23,220

Per month = $23,220 / 12 = $1,935

Hope you found this helpful , good luck !

7 0

3 years ago

Read 2 more answers

Researchers are investigating the effectiveness of leg-strength training on cycling performance. A sample of 7 men will be selec

ruslelena [56]

Answer:

A. The interval will be narrower if 15 men are used in the sample.

Step-by-step explanation:

Hello!

When all other things remain the same, which of the following statements about the width of the interval is correct?

A. The interval will be narrower if 15 men are used in the sample.

B. The interval will be wider if 15 men are used in the sample.

C. The interval will be narrower if 5 men are used in the sample.

D. The interval will be narrower if the level is increased to 99% confidence.

E. The interval will be wider if the level is decreased to 90% confidence.

Consider that the variable of interest "Xd: Difference between the peak power of a cyclist before training and after training" has a normal distribution. To construct the confidence interval for the population mean of the difference you have to use a pooled t-test.

The general structure for the CI is "point estimate"±" margin fo error"

Any modification to the sample size, sample variance and/or the confidence level affect the length of the interval (amplitude) and the margin of error (semiamplitude)

The margin of error of the interval is:

d= $t_{n-1;1-\alpha /2}$ * (Sd/n)

1) The sample size changes, all other terms of the interval stay the same.

As you can see the margin of error and the sample size (n) have an indirect relationship. This means, that when the sample size increases, the semiamplitude decreases, and when the sample size decreases, the semiamplitude increases.

↓d= $t_{n-1;1-\alpha /2}$ * (Sd/↑n)

↑d= $t_{n-1;1-\alpha /2}$ * (Sd/↓n)

Correct option: A. The interval will be narrower if 15 men are used in the sample.

2) The confidence level has a direct relationship with the semiamplitude of the interval, this means that when the confidence level increases, so do the semiamplitude, and if the level decreases, so do the semiamplitude:

↓d= ↓ $t_{n-1;1-\alpha /2}$ * (Sd/n)

↑d= ↑ $t_{n-1;1-\alpha /2}$ * (Sd/n)

I hope it helps!