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Nikolay [14]
4 years ago
10

Simplify: (-2a^4b) • (5a^4b^3)

Mathematics
2 answers:
lisov135 [29]4 years ago
4 0

Answer:

(-2a^4b) x (5a^4b^3) simplified :

-10a^4b+4b^3

Hope this helps,

Davinia.

Aleksandr [31]4 years ago
3 0

Answer:

(-10a^{4b+4b^{3}})

Step-by-step explanation:

Given in the question the expression,

(-2a^{4b} )(5a^{4b^{3}})

Step1

Product Rule

The exponent "product rule" tells us that, when multiplying two powers that have the same base, you can add the exponents.

so,

(-2a^{4b})(5a^{4b^{3} })=(-10a^{4b+4b^{3}})

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3 years ago
The relationship between the standard normal random variable z and normal random variable X is that :
alisha [4.7K]

Answer:

(B) The standard normal variable Z counts the number of standard deviations that the value of the normal random variable X is away from its mean

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Positive z-score: Above the mean

Negative z-score: Below the mean

All variables are continuous.

X can be positive or negative, just like Z

So the correct answer is:

(B) The standard normal variable Z counts the number of standard deviations that the value of the normal random variable X is away from its mean

4 0
3 years ago
How many words can be formed by using the W,X,Y,Z if repetitions is not allowed?
Setler [38]

Answer:

24

Step-by-step explanation:

What you have here is a permutation, seeing as each element can only be used once.

We have 4 letters initially, so we can choose any 1 as our first letter. We have 4 choices for our first letter

However, once we choose our first letter, we can't use it anymore, so, for our second letter, we can only choose from the remaining 3 letters.

Furthermore, once we choose our second letter, we can only choose our 3rd letter from the remaining two letters we didn't choose yet.

Finally, our last letter will always be the one we didn't choose the last 3 times. So there is only one choice here.

Going off of this, we have four choices for the 1st letter, three choices for the 2nd letter, two choices for the 3rd letter, and one choice for the 4th letter

The way to calculate how many permutations we have without repetition is using factorials

N!

Where N is the number of elements you have.

In this case, it would be 4!

4! is 4 * 3 * 2 * 1

Which equals 24

If you notice, each number in 4! is the number of options we have for each choice. 4, then 3, and so on

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