Can someone please help I am stuck on this question

Find the values for a and b that would make the following equation true (ax^2)(-6x^b)=12x^5

dedylja [7]

Answer:

x^b=x^3

Step-by-step explanation:

(ax^2)(-6x^b)=12x^5

(a)(6)=6a or a6

(x^2)(x^b)=x^5

x^3 because

(a^b)(a^c)=a^b+c

Hope this helped.

3 0

2 years ago

2x - 5y +10x - 2y + 3z solve.

Margarita [4]

Answer:

12x-7y+3z

Step-by-step explanation:

8 0

2 years ago

If the functions y=9/2x^2 Was placed in the form y=ax^b , where a and b are real numbers, then which of the following is the val

exis [7]

Answer:

$\frac{13}{2}$

Step-by-step explanation:

We have that:

$a=\frac{9}{2} \\b=2$

Now we can find a+b

$\frac{9}{2}+2=\frac{9+4}{2}=\frac{13}{2}$

5 0

2 years ago

A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that

ZanzabumX [31]

Answer:

93.32% probability that a randomly selected score will be greater than 63.7.

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:

$\mu = 80.2, \sigma = 11$

What is the probability that a randomly selected score will be greater than 63.7.

This is 1 subtracted by the pvalue of Z when X = 63.7. So

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

$Z = \frac{63.7 - 80.2}{11}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

1 - 0.0668 = 0.9332

93.32% probability that a randomly selected score will be greater than 63.7.

5 0

3 years ago

Different dealers may sell the same car for different prices. The sale prices for a particular car are normally distributed with

Stels [109]

Answer:

P(X > 25) = 0.69

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be 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.

The sale prices for a particular car are normally distributed with a mean and standard deviation of 26 thousand dollars and 2 thousand dollars, respectively.

This means that $\mu = 26, \sigma = 2$

Find P(X>25)

This is 1 subtracted by the pvalue of Z when X = 25. So

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

$Z = \frac{25 - 26}{2}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.31

1 - 0.31 = 0.69.

So

P(X > 25) = 0.69

4 0

2 years ago

Can someone please help I am stuck on this question​

Can someone please help I am stuck on this question