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3 years ago

What is 6 times (12-4) in words?

Mathematics

2 answers:

kotegsom [21]3 years ago

8 0

Twelve munis four times six

Alika [10]3 years ago

3 0

Six times twelve minus 4 equals forty-eight

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How the Right of Occupancy can be transferred or exchanged.

Radda [10]

It can be transferred

8 0

3 years ago

Suppose an automotive repair company wants to determine the current percentage of customers who keep up with regular vehicle mai

Nina [5.8K]

Answer: 1692

Step-by-step explanation:

Formula to find the sample size :

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

Given : Confidence level : $(1-\alpha)=0.90$

⇒ significance level = $\alpha= 0.10$

z-value for 90% confidence interval (using z-table)= $z_{\alpha/2}=1.645$

Prior estimate of the population proportion (p) of customers who keep up with regular vehicle maintenance is unknown.

Let we take p= 0.5

Margin of error : E= 2%=0.02

Now, the required sample size will be :

$n=0.5(1-0.5)(\dfrac{1.645}{0.02})^2$

Simplify , we get

$n=(0.25)(6765.0625)=1691.265625\approx1692$

Hence, the required sample size = 1692

8 0

4 years ago

Read 2 more answers

which expression is equivalent to -28x^2+35x ? F. 7x(4x+5) G.-7x(4x-5) H.7x(4x-5) J.-7x(4x+5) please explain w the steps

Ludmilka [50]

Hey!!

Solution,

-28x^2+35x

take the common,

-7x(4x-5)

The answer is -7x(4x-5)

Here,

If we solve,

-7x(4x-5)

Then, the answer would be:

= -7x*4x+7x*5

= -28x^2+35x

Hope it helps

Good luck on your assignment

4 0

3 years ago

Read 2 more answers

Use the equation for variance below, along with the given data set, to answer the following questions.

juin [17]

nx -33=9x-4

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A tire manufacturer knows that 5% of tires contain a defect, and the presence of a defect is independent from tire to tire.

saul85 [17]

The probability that if 5 tires are inspected, exactly 2 have a defect is 0.021

<h3>How to calculate the probability</h3>

The given parameters are:

$p = 5\%$ --- the proportion that contains defect

$n =5$ --- the sample size

$r =2$ -- the selected sample

The probability is then calculated as:

$P(x =r) = ^nC_r * p^r * (1 - p)^{n-r}$

So, we have:

$P(x =2) = ^5C_2 * (5\%)^2 * (1 - 5\%)^{5-2}$

$P(x =2) = 10 * (5\%)^2 * (1 - 5\%)^3$

$P(x =2) = 0.021$

Hence, the probability that if 5 tires are inspected, exactly 2 have a defect is 0.021