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2 years ago

Simplify 6m2n3 – 3m3n2 + 8m2n3 + 12m3n2

1 answer:

6 0

Hello! Hope I can assist you.

Let us simplify some terms first!

$6m\cdot \:2n\cdot \:3 \ \textgreater \ 36mn$
$3m\cdot \:3n\cdot \:2 \ \textgreater \ 2\cdot \:3^2mn$
$8m\cdot \:2n\cdot \:3 \ \textgreater \ 48mn$
$12m\cdot \:3n\cdot \:2 \ \textgreater \ 72mn$

$36mn-2\cdot \:3^2mn+48mn+72mn$

From here, group like terms
$36mn-2\cdot \:3^2mn+48mn+72mn$

$\mathrm{Add\:similar\:elements:}\:36mn+48mn+72mn=156mn$
$156mn-2\cdot \:3^2mn$

$2\cdot \:3^2mn \ \textgreater \ 2\cdot \:9mn \ \textgreater \ 18mn$

$156mn-18mn$

$\mathrm{Add\:similar\:elements:}\:156mn-18mn=138mn$

Hope this helps!

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7. Reduce the fraction 36/48 to its lowest terms. A. 1 12/48 B. 3/4 C. 1 1/4 D. 6/8

Pie

The answer is B. 3/4

36/48 both share the common factor of 12, divide 12 by the numerator and denominator to get the simplified fraction, 3/4

3 0

3 years ago

Read 2 more answers

I will mark you brainliest for the right answer and you must HURRY to get it! 15 points!

DENIUS [597]

A triangles three angles always sum up to 180 degrees. We must find angle B.

We already have angle A and C, so it's quite simple.

Angle A is 28*, and angle C is 36*

We add these two together

28 + 36 = 64

Now we subtract

180 - 64 = 116

Angle B = 116*

Hope this helps you! (:

-Hamilton1757

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2 years ago

Question in Picture.

almond37 [142]

Answer:

The correct answer would be C

Step-by-step explanation:

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2 years ago

Solve the system of equations by graphing on your own paper. What is the y

xenn [34]

Answer: The coordinate of the solution is (0.25,-0.5)

Step-by-step explanation:

the solution is (0.25,-0.5) because when you graph them on a graph, they both intercept at that point.

3 0

2 years ago

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65

erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 2.65

$\sigma$ = standard deviation = 0.85

n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

P(X bar <= 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z < 2.06) = 0.98030

P(X bar <= 2.65) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .

8 0

3 years ago