How is 225 divisible by nine

2 answers:

8 0

<span>A number is divisible by 9 if the sum of the digits is divisible by 9.

</span><span>225 is divisible by 9 since the sum of the digits is 9 , (2+2+5=19), and 9 is divisible by 9.</span>

Ivanshal [37]3 years ago

5 0

A number is divisible by 9 if and only if the sum of its number is a multiple of nine.

2+2+5=9 hence 255 is divisible by 9.

You might be interested in

A marketing research company needs to estimate the average total compensation of CEOs in the service industry. Data were randoml

mezya [45]

SOLUTION:

Step 1 :

In this question, we are told that a marketing research company needs to estimate the average total compensation of CEOs in the service industry.

We also have that: Data were randomly collected from 38 CEOs and the 98% confidence interval was calculated to be ($2,181,260, $5,836,180).

Then, we are asked to find the margin error for the confidence interval.

Step 2:

We need to recall that:

$\text{Higher Confidence Interval, CI}_{H\text{ = }}X\text{ + }\frac{Z\sigma}{\sqrt[]{n}}$ $\text{Lower Confidence Interval , CI}_{L\text{ }}=\text{ X - }\frac{Z\sigma}{\sqrt[]{n}}$

It means that:

$\vec{}X\text{ = }\frac{CI_{H\text{ }}+CI_L}{2}$ $\text{Margin of error, }\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{CI_{H\text{ - }}CI_L}{2}$

where,

$CI_H\text{ = }$$5,836,180$\text{ }$$$$ $CI_{L\text{ }}=\text{ }2,181,260$

putting the values into the equation for the margin of error, we have that:

$\text{Margin of error,}\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{5,836,180\text{ - }2,181,260\text{ }}{2}$ $\begin{gathered} =\text{ }\frac{3654920}{2} \\ =1,\text{ 827, 460} \end{gathered}$

CONCLUSION:

The margin error for the confidence interval is 1, 827, 460

5 0

10 months ago

B=(2x+3)(4x^2-6x+9)-2(4^3-1)

nexus9112 [7]

B = (2x+3)(4x^2-6x+9)-2(4^3-1)
B = 8x^3-99

Hope it helps : )

5 0

1 year ago

What is the answer of 2×82

slamgirl [31]

The answer to this problem is 164.

Hope this answer helps! feel free to ask any additional questions :)