Identify the terms and the like terms in this expression.

What's the absolute value of -|6| + 6? Options : O. 0 O. 12 O. -12

vekshin1

Answer:

Zero

Step-by-step explanation:

any absolute value bars come off so -6+6=0

3 0

3 years ago

Read 2 more answers

Which of the following is an example of the compliment rule being applied to

UNO [17]

Answer:

Hello There!!

Step-by-step explanation:

I think the answer is B. The probability of not drawing a club or jack from a deck of cards.

hope this helps,have a great day!!

~Pinky~

6 0

3 years ago

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Use the distributive property to remove the parenthesis (v+12)10

serg [7]

The distributive property means to distribute the number being multiplied by the numbers in the parentheses.
In this case, (v+12) is being multiplied by 10.
To solve this, you simply have to multiply 10 by each term inside the parentheses.
That means it would be (v•10)+(12•10).
(v•10) is equal to 10v, since v is the variable and we don't know its value.
(12•10) is equal to 120.
So by using the distributive property we can find that (v+12)10 = 10v+120.

6 0

3 years ago

The number of violent crimes committed in a day possesses a distribution with a mean of 2.2 crimes per day and a standard deviat

VARVARA [1.3K]

Answer:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And replacing we got:

$\mu_{\bar X}= 2.2$

$\sigma_{\bar X}= \frac{6}{\sqrt{100}}= 0.6$

Step-by-step explanation:

For this case we have the following info given:

$\mu = 2.2$ represent the mean

$\sigma = 6$ represent the deviation

We select a sample size of n=100. This sample is >30 so then we can use the central limit theorem. And we want to find the distribution for the sample mean and we know that the distribution is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And replacing we got:

$\mu_{\bar X}= 2.2$

$\sigma_{\bar X}= \frac{6}{\sqrt{100}}= 0.6$

6 0

4 years ago

A data set includes weights of garbage discarded in one week from 62 different households. The paired weights of paper and glass

lara [203]

Answer:

Because the absolute value of the test statistic is less than the positive critical value, there is not enough evidence to support the claim that there is a linear correlation between the weights of discarded paper and glass for a significance level of α = 0.05.

Step-by-step explanation:

The correlation matrix provided is:

Variables Paper Glass

Paper 1 0.1853

Glass 0.1853 1

Te hypothesis for the test is:

H₀: ρ = 0 vs. H₀: ρ ≠ 0

The test statistic is:

r = 0.1853 ≈ 0.185

As the alternate hypothesis does not specifies the direction of the test, the test is two tailed.

The critical value for the two-tailed test is:

$r_{\alpha/2, (n-2)}=r_{0.05/2, (62-2)}=r_{0.05/2, 60}=0.250$

The conclusion is:

Because the absolute value of the test statistic is less than the positive critical value, there is not enough evidence to support the claim that there is a linear correlation between the weights of discarded paper and glass for a significance level of α = 0.05.

6 0

4 years ago