Simplify the rational expression: (5a^2b^2c^-3)/(2a^3)

1 answer:

7 0

First we can move c^-3 to the denominator as c^3 since it has a negative exponent.

(5a^2*b^2)/(2a^3*c^3)

Now we can reduce the expression. We have 'a' terms on top and bottom, so to reduce we just subtract the exponents. We have a^2 on top and a^3 on bottom, so we are left with a^1 on bottom.

(5b^2)/(2a*c^3)

You might be interested in

Please help me with this....

amm1812

The answer to the question is a

4 0

4 years ago

X + y = 6

sergejj [24]

X=14, is the answer I believe. Good luck!

6 0

3 years ago

suppose you are playing a game of chance. If you bet $5 on a certain event, you will collect $115 (including your $5 bet) if you

malfutka [58]

1) It is 1:27 (you pay $1 and if the horse wins you get $28 and if he loses you lose the $1.

<span>If the odds are 1:1 it represents a 50% chance and in that case you would pay $1 and get back either $0 o</span><span>. </span>

7 0

3 years ago

Read 2 more answers

Kendrick received 24 fewer e-mails than he did on Thursday. If he received a total of 126 e-mails on those two days, how many di

telo118 [61]

E+(e-24)=126

2e-24=126

2e=150

e=75

So (s)he recieved 75 and 75-24=51 emails on those days.

3 0

4 years ago

(t-distribution) A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. To m

statuscvo [17]

Answer:

There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.

Step-by-step explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. This means that $\mu = 50$ .

We found that the sample had a average lifespan of 47.3 months, and a standard deviation of s = 9 months. What is the probability that we find this lifespan for our sample average, or something even shorter?

We have to find the pvalue of Z when $X = 47.3$ .