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3 years ago

Simplify 3a^2b^2x2ab+6a^3b^2

Mathematics

1 answer:

ira [324]3 years ago

3 0

$3 {a}^{2} \times {b}^{2} \times 2ab + 6 {a}^{3} \times {b}^{2} =$

$6 {a}^{3} {b}^{3} + 6 {a}^{3} {b}^{2} =$

Factoring 6a^3 b^2

$6 {a}^{3} {b}^{2}(b + 1)$

_________________________________

I think this is the correct answer.

And we're done.

Thanks for watching buddy good luck.

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Find the 75th term of the following6, 10, 14, 18,

enot [183]

Given the sequence:

6, 10, 14, 18,...

We will find the 75th term

The given sequence is an arithmetic sequence

Because there is a constant common differnce

d = 18 - 14 = 14 - 10 = 10 - 6 = 4

The first term = a = 6

The general formula of the arithmetic sequence is as follows:

$a_n=a+d(n-1)$

Where: n is the nth term

To find the 75th term, substitute with n = 75 and a = 6, d = 4

$a_{75}=6+4\cdot(75-1)=302$

So, the answer will be the 75th term = 302

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11 months ago

How to do exponents with variable

storchak [24]

Answer:

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Step-by-step explanation:

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2 years ago

Devora explored a secret cave. 48\text{ m}48 m48, start text, space, m, end text from the entrance, she found an empty chest wit

siniylev [52]

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3 0

3 years ago

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

3 years ago

Can someone help me, I don’t get this

dmitriy555 [2]

I'll break it down for you. Here are the main points:

-has 2 large servings

-made from 9 pints

-x represents how many pints

-y represents how many servings

Now that we know this we can find what each serving comes out to be for example:

$\frac{9}{2\\}$ =4.5 or 4 $\frac{1}{2}$

So for each serving it will come out to be 4.5 pints.

Now that we know this we can figure out the slope to find what the equation which is for example the slope for this equation would be rise over run for example:

$\frac{rise}{run}$ = $\frac{1}{4.5}$ = $\frac{2}{9}$ when simplified.

So our equation would be (also the answer) y= $\frac{2}{9}$ x

Hope this helps my dude ^^ If you have more questions go ahead and ask.

5 0

3 years ago