The sum of their speeds is 380 miles per hour. The difference of their speeds is 30 miles per hour. Then the faster plane is traveling at
.. (380 +30)/2 = 205 miles per hour

The slower plane's speed is 30 miles per hour less, so it is traveling at
.. 205 mph -30 mph = 175 miles per hour

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=y%3Dmx%2Bb%20for%20b" id="TexFormula1" title="y=mx+b for b" alt="y=mx+b for b" align="absmiddl

Gemiola [76]

Minus mx from both sides
y-mx=b
b=y-mx

7 0

3 years ago

83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

$\chi^2_{\alpha/2}=104.139$

$\chi^2_{1- \alpha/2}=62.132$

The confidence interval is given by:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

5 0

3 years ago