Explain your answer

Factor each expression and explain how you solved it.

rosijanka [135]

42. g^2+2g-7g-14= g(g+2)-7(g+2) = (g-2)(g+7)
44. 6k^2-6kL-4kL+4L^2= 6k(k-L)-4L(k-L)= (6k-4L)(k-L)
46. r^2-4r+10r-40 = r(r-4)+10(r-4)= (r+10)(r-4)
48. t^2-3t-10t-30 = t(t-3)-10(t-3)= (t-10)(t-3)
50. 3x^2-3x+6x-6 = 3x(x-1)+6(x-1) = (3x+6)(x-1)

REALLY hoping this helps bc this took me some time

7 0

3 years ago

Suppose that the researchers wanted to estimate the mean reaction time to within 6 msec with 95% confidence. Using the sample st

Lisa [10]

Answer:

a) The 95% confidence interval would be given by (509.592;550.308)

b) n=523 rounded up to the nearest integer

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=530$ represent the sample mean for the sample

$\mu$ population mean

s=70 represent the sample standard deviation

n=48 represent the sample size (variable of interest)

Confidence =95% or 0.995

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are df=n-1=48-1=47

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,47)".And we see that $z_{\alpha/2}=2.01$