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2 years ago

Is the LCM of a pair of numbers ever equal to one of the numbers?

Mathematics

1 answer:

Nata [24]2 years ago

8 0

Yes . It happens depending on the numbers!

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Michael bought 3/4 pound of turkey. He ate 1/3 of the turkey. How much turkey did Michael eat?

Lana71 [14]

The answer is b
hope this helps

7 0

2 years ago

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There are 32 sixth graders and 44 seventh graders in the video gaming club. Which ratio is the greatest?

katrin [286]

I’m gonna say D hope this helps

3 0

2 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

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A random sample of 100 observations from a quantitative population produced a sample mean of 22.8 and a sample standard deviatio

natima [27]

Answer:

We conclude that the population mean is 24.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 24

Sample mean, $\bar{x}$ = 22.8

Sample size, n = 100

Alpha, α = 0.05

Sample standard deviation, s = 8.3

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 24\\H_A: \mu \neq 24$

We use Two-tailed z test to perform this hypothesis.

Formula:

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

Putting all the values, we have

$z_{stat} = \displaystyle\frac{22.8 - 24}{\frac{8.3}{\sqrt{100}} } = -1.44$

We calculate the p-value with the help of standard z table.

P-value = 0.1498

Since the p-value is greater than the significance level, we accept the null hypothesis. The population mean is 24.

Now, $z_{critical} \text{ at 0.05 level of significance } = \pm 1.96$

Since, the z-statistic lies in the acceptance region which is from -1.96 to +1.96, we accept the null hypothesis and conclude that the population mean is 24.

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2 years ago

Torrey has $20 to buy a birthday present for her that she does this to buy a CD for $18 the sales tax is 7% does she have enough

Effectus [21]

No, because she will go over by .57 cents. So she won't be able to buy the gift. You can find the answer by dividing 18 by 7 and adding that answer to 18 to get the amount total you owe including tax.

8 0

3 years ago