Vehicle Car A is now passing a truck that is 90 feet long. Vehicle A is traveling 40 mph (or 60 feet per second) and

An SRS of 350 high school seniors gained an average of x¯=21 points in their second attempt at the SAT Mathematics exam. Assume

Lynna [10]

Answer:

a) $21-2.58\frac{52}{\sqrt{350}}=13.83$

$21+2.58\frac{52}{\sqrt{350}}=28.17$

So on this case the 99% confidence interval would be given by (13.83;28.17)

b) $ME=2.58\frac{52}{\sqrt{350}}=7.17$

c) $ME=2.58\frac{52}{\sqrt{100}}=13.42$

d) D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

Step-by-step explanation:

Previous concepts

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".

Part a

$\bar X=21$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=52$ represent the population standard deviation

n=350 represent the sample size

99% confidence interval

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

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

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$21-2.58\frac{52}{\sqrt{350}}=13.83$

$21+2.58\frac{52}{\sqrt{350}}=28.17$

So on this case the 99% confidence interval would be given by (13.83;28.17)

Part b

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{350}}=7.17$

Part c

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{100}}=13.42$

Part d

As we can see when we reduce the sample size we increase the margin of error so the best option for this case is:

D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

6 0

2 years ago

If x+3 is a factor of x^3-x^2-10x+A. What is the value of A?

Leno4ka [110]

Answer:

$A = 6$ .

Step-by-step explanation:

Let $a$ denote a constant (a number, rather than a variable like $x$ .)

Let $f(x)$ denote an expression about $x$ (for example,

By the factor theorem, if $(x - a)$ is a factor of $f(x)$ , then $f(a) = 0$ . In other words, if all mentions of $x$ in the expression $f(x)\!$ are replaced with $a$ , then the expression should evaluate to zero.