All categories

2 years ago

NEED HELP! Thanks!

Mathematics

1 answer:

sladkih [1.3K]2 years ago

6 0

I think the answer would be 7.5 hours or 4 hours,not so sure which one exactly

You might be interested in

What number has 7 ones, 3 fewer thousands than ones, 5 fewer hundred thousands than ones, 2 more tens than hundred thousands, 2

Harman [31]

Answer:

Step by step below

Step-by-step explanation:

0 000 007

0 004 007

0 204 007

0 204 047

0 204 647

0 224 647

5 0

1 year ago

The low temperatures for 5 days was 5, -2, -11 ,0 and-3.What was the average low temperature for those days

Daniel [21]

Data:
Average (A): ?
Arithmetic mean (A.M) = 5, -2, -11, 0, -3sum of n numbers = 5

Solving:
$A = \frac{AM}{n} = \frac{5+(-2)+(-11)+0+(-3)}{5} = \frac{5-2-11-3}{5} = \frac{\diagup\!\!\!\!3-11-\diagup\!\!\!\!3}{5} = \frac{-11}{5} = - 2.2$

Answer:

<span>The average low temperature of these days is -2.2</span>

3 0

2 years ago

The system has infinitely many solutions. Rearranging the left side of the 2nd equation and subtracting 3a from both sides gives

Alex_Xolod [135]

Answer:

https://web.williams.edu/Mathematics/sjmiller/public_html/105/hwsolns/HWSolns_Math105_Sp2013.pdf click the link

Step-by-step explanation:

5 0

2 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.