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3 years ago

Which expression is equivalent to (image attached)

Mathematics

2 answers:

Monica [59]3 years ago

5 0

<h2>Hello!</h2>

The answer is:

B. $x(7x^{2}-3x+9)-3(7x^{2}-3x+9)$

We are given a simplified expression and we are asked to know which of the given options is equivalent to the simplified expression, so, we need to apply the distributive property in order to find the correct choice.

The distributive property states that:

$(a+b)(c+d+e)=a(c+d+e)+b(c+d+e)=ac+ad+ae+bc+bd+be$

So, if we are given the following expression:

$(x-3)(7x^{2}-3x+9)$

Then, we need to apply the distributive property in order to find the equivalent expression:

$(x-3)(7x^{2}-3x+9)=x(7x^{2}-3x+9)-3(7x^{2}-3x+9)$

Hence, the correct option is:

B. $x(7x^{2}-3x+9)-3(7x^{2}-3x+9)$

Have a nice day!

Roman55 [17]3 years ago

4 0

Answer:

B. $x(7x^2-3x+9)-3(7x^2-3x+9)$

Step-by-step explanation:

The given expression is $(x-3)(7x^2-3x+9)$

Recall and use the distributive property;

In this case;

$(a-b)(c+d+e)=a(c+d+e)-b(c+d+e)$

This implies that;

$(x-3)(7x^2-3x+9)=x(7x^2-3x+9)-3(7x^2-3x+9)$

Therefore the correct choice is B.

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2 years ago

9. Gold miners in Alaska have found, on average, 12 ounces of gold per 1000 tons of dirt excavated with a standard deviation of

netineya [11]

Answer:

9.18% probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated

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.

In this problem, we have that:

$\mu = 12, \sigma = 3$