The LCM of 2 numbers is 60 and 1 of the numbers is 7 less than the other number. What are the numbers?

1 answer:

5 0

Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.

Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:

discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)

and

-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)

Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.

Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.

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d. 68.39%.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 99% confidence level, hence $\alpha = 0.99$ , z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$ , so the critical value is z = 2.575.