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3 years ago

Need help with theses questions please

Mathematics

1 answer:

Olin [163]3 years ago

7 0

Hello:
3x²+5x+2=0
delta = (5)²-4(3)(2)=1
x =( -5 ±1)/6
x= -2/3 or x=-1

t4-81=(t²)² -9² = (t²-9)(t²+9) =(t-3)(t+3)(t²+9)....(answer B)

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I need help with multistep equations

Over [174]

Answer:

1 2/3 = m

Step-by-step explanation:

2/3 = m+3/5 -8/5

Combine terms

2/3 = m-5/5

2/3 = m -1

Add 1 to each side

2/3 +1 = m-1+1

2/3 +3/3 = m

5/3 = m

1 2/3 = m

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3 years ago

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What is the value of z?

OLga [1]

Answer:

z = 86

Explanation:

90 - 68 = 12
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3 years ago

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3 years ago

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:

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

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

8 0

3 years ago