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3 years ago

10q-30 is equivalent to what ?

Mathematics

1 answer:

butalik [34]3 years ago

3 0

Answer:

10(q-3)

last option

please brainliest me btw

Step-by-step explanation:

The expression states 10q-30

The GCF is 10 (30 GCF by 10 Is 10)

Factor out 10

After, divide all the terms by 10

10 (10q/10) - 30/10

Simplify this and we get

10(q-3)

so in conclusion, it is the last option.

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Sloan [31]

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3 years ago

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Transform x2 + 6x + 8 = 0 into the form (x − p)2 = q?

liraira [26]

So hmmm x²+6x+8=0

alrite.. let's do some grouping now
( x² + 6x + [?]²) + 8 = 0

notice above, we have a missing fellow in order to get a perfect square trinomial... hmm who would that be?

let's take a peek at the middle guy of the trinomial.. 6x.. hmmm let's factor it, 2*3*x, wait a minute! 2 * 3 * x... we already have x² on the left-side, since the middle term is just 2 * the square root of the other two terms, that means that the guy on the right, our missing guy must be "3"

alrite, let's add 3² then, however, bear in mind that, all we're doing is borrowing from our very good friend Mr Zero, 0

so if we add 3², we also have to subtract 3², let's do so

(x² + 6x +3² - 3²) + 8 = 0

(x² + 6x +3²) + 8 - 3² = 0

(x+3)²=3² - 8

(x+3)² = 1

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3 years ago

Read 2 more answers

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

mojhsa [17]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, MOE = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, MOE = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$