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3 years ago

Factor the following expression using the GCF. 4dr – 16r r

Mathematics

2 answers:

AveGali [126]3 years ago

8 0

The greatest common factor is 4r so

4r(d-4)

xxMikexx [17]3 years ago

7 0

4dr-16r

Remember, the first step in doing this is finding the GCF, or greatest common factor.

Look at both equations, and see what it is.

We see that it is 4.

Look for any variables that are found in both numbers.

r is found in both numbers, so 4r is the GCF.

4dr-16r

What is 4r divided by 4dr? It would be d.

What is -16r divided by 4r? -4

So, our equation now looks like this:

4r(d-4)

So, the last one is correct.

~Hope I helped!~

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Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of

jarptica [38.1K]

Answer:

The "probability that a given score is less than negative 0.84" is $\\ P(z$ .

Step-by-step explanation:

From the question, we have:

The random variable is normally distributed according to a standard normal distribution, that is, a normal distribution with $\\ \mu = 0$ and $\\ \sigma = 1$ .
We are provided with a z-score of -0.84 or $\\ z = -0.84$ .

Preliminaries

A z-score is a standardized value, i.e., one that we can obtain using the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

x is the raw value coming from a normal distribution that we want to standardize.
And we already know that $\\ \mu$ and $\\ \sigma$ are the mean and the standard deviation, respectively, of the normal distribution.

A z-score represents the distance from $\\ \mu$ in standard deviations units. When the value for z is negative, it "tells us" that the raw score is below $\\ \mu$ . Conversely, when the z-score is positive, the standardized raw score, x, is above the mean, $\\ \mu$ .

Solving the question

We already know that $\\ z = -0.84$ or that the standardized value for a raw score, x, is below $\\ \mu$ in 0.84 standard deviations.

The values for probabilities of the standard normal distribution are tabulated in the standard normal table, which is available in Statistics books or on the Internet and is generally in cumulative probabilities from negative infinity, - $\\ \infty$ , to the z-score of interest.

Well, to solve the question, we need to consult the standard normal table for $\\ z = -0.84$ . For this:

Find the cumulative standard normal table.
In the first column of the table, use -0.8 as an entry.
Then, using the first row of the table, find -0.04 (which determines the second decimal place for the z-score.)
The intersection of these two numbers "gives us" the cumulative probability for z or $\\ P(z$ .

Therefore, we obtain $\\ P(z$ for this z-score, or a slightly more than 20% (20.045%) for the "probability that a given score is less than negative 0.84".

This represent the area under the standard normal distribution, $\\ N(0,1)$ , at the left of z = -0.84.

To "draw a sketch of the region", we need to draw a normal distribution (symmetrical bell-shaped distribution), with mean that equals 0 at the middle of the distribution, $\\ \mu = 0$ , and a standard deviation that equals 1, $\\ \sigma = 1$ .

Then, divide the abscissas axis (horizontal axis) into equal parts of one standard deviation from the mean to the left (negative z-scores), and from the mean to the right (positive z-scores).

Find the place where z = -0.84 (i.e, below the mean and near to negative one standard deviation, $\\ -\sigma$ , from it). All the area to the left of this value must be shaded because it represents $\\ P(z$ and that is it.

The below graph shows the shaded area (in blue) for $\\ P(z$ for $\\ N(0,1)$ .

7 0

3 years ago

Work Out 8^1/3 ÷ 5^-2

svp [43]

I believe the answer would be 50

explanation:

calculate
2 ÷ 5^-2

use negative power rule x^-a = 1/x^a
2 ÷ 1/5^2

simply 5^2 to 25
2 ÷ 1/25

use this rule a ÷ b/c = a • c/b
2 • 25

simplify
50

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Phantasy [73]

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3 years ago

Need the answer asap i will give you brainliest

ladessa [460]

Answer:

Step-by-step explanation:

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2 years ago

In proving statements using the coordinate method, the placement of the figure can simplify the algebraic work.

Salsk061 [2.6K]

Answer: The correct answer is True.

When you are making coordinate proofs, you are using the value of the coordinates in your proof and your algebra work.

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