To eliminate x, you can multiply the second equation by 2. This will make -2x into -4x, so that 4x and 4x will be able to cancel out leaving you to solve for y.

To eliminate y, you can multiply the second equation by 3. This will make 3y into 9y, so that -9y and 9y will be able to cancel out leaving you to solve for x.
Hope this helped.

3 0

2 years ago

In a large lecture class, the professor announced that the scores on a recent exam were normally distributed with a range from 5

iVinArrow [24]

Answer:

The standard deviation of the scores on a recent exam is 6.

The sample size required is 25.

Step-by-step explanation:

Let X = scores of students on a recent exam.

It is provided that the random variable X is normally distributed.

According to the Empirical rule, 99.7% of the normal distribution is contained in the range, μ ± 3σ.

That is, P (μ - 3σ < X < μ + 3σ) = 0.997.

It is provided that the scores on a recent exam were normally distributed with a range from 51 to 87.

This implies that:

P (51 < X < 87) = 0.997

So,

μ - 3σ = 51...(i)

μ + 3σ = 87...(ii)

Subtract (i) and (ii) to compute the value of σ as follows:

μ - 3σ = 51

(-)μ + (-)3σ = (-)87

______________

-6σ = -36

σ = 6

Thus, the standard deviation of the scores on a recent exam is 6.

The (1 - α)% confidence interval for population mean is given by:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The margin of error of this interval is:

$MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

Given:

MOE = 2

σ = 6

Confidence level = 90%

Compute the z-score for 90% confidence level as follows:

$z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645$

*Use a z-table.

Compute the sample required as follows:

$MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\2=1.645\times \frac{6}{\sqrt{n}}\\n=(\frac{1.645\times 6}{2})^{2}\\n=24.354225\\n\approx 25$

Thus, the sample size required is 25.

4 0

3 years ago