Which of the following is the algebraic expression that best describes the height y, if x is the width?

Find the length of segment AB.

12345 [234]

The answer is D.
explain:
A^2 + b^2 = c^2

8 0

3 years ago

Read 2 more answers

2 ( t-3) > 2t-8, what is the correct answer

liq [111]

$2(t-3) > 2t-8\ \ \ \ |\text{use distributive property}\\\\(2)(t)+(2)(-3) > 2t-8\\\\2t-6 > 2t-8\ \ \ \ \ |\text{subtract 2t from both sides}\\\\-6 > -8\ \ \ TRUE\\\\Answer:\ t\in\mathbb{R}\ (all\ real\ numbers)$

8 0

3 years ago

Required information An article presents a new method for timing traffic signals in heavily traveled intersections. The effectiv

Natasha2012 [34]

Answer:

$652.6-2.01\frac{311.7}{\sqrt{50}}=563.997$

$652.6+2.01\frac{311.7}{\sqrt{50}}=741.203$

So on this case the 95% confidence interval would be given by (563.997;741.203)

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".

$\bar X=652.6$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=311.7 represent the sample standard deviation

n=50 represent the sample size

Soltuion to the problem

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

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=50-1=49$

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(0.025,49)".And we see that $t_{\alpha/2}=2.01$