Subtract 2/5 from the product of 3.5 and 4.2

Rovan can make 48 tarts per hour. complete the table below:

GarryVolchara [31]

Answer:

A.) in 6 hours Rovan can make 288 tarts

B. at this rate it'll take Rovan about 2 and a half hours to make 120 tarts

Step-by-step explanation:

so if Rovan can make 48 tarts an 1 hour take 48 and multiply it by the hours to find out how many tarts he can make

48 x 1 =48

48 x 2 =96

48 x 3 =144 .....

48 x 6 =288

A.) in 6 hours Rovan can make 288 tarts

B. at this rate it'll take Rovan about 2 and a half hours to make 120 tarts

3 0

2 years ago

Read 2 more answers

Solve for q. mn + q = p q = p - mn q = p (mn ) q = p + mn

cupoosta [38]

It would be your first answer because since the MN is a positive you have to subtract it to move it to P and you solved for Q

6 0

3 years ago

Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the

ziro4ka [17]

Answer:

a) $\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

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

$19.08 \leq \sigma \leq 66.56$

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

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

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

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=9-1=8$

The Confidence interval is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and the critical values are:

$\chi^2_{\alpha/2}=20.09$

$\chi^2_{1- \alpha/2}=1.65$

And replacing into the formula for the interval we got:

$\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

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

$19.08 \leq \sigma \leq 66.56$

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0

3 years ago

There are two sizes of trucks available: larger truck with a 2.5-ton payload, and smaller trucks with 1.5-ton payload. To unload

amid [387]

You would need 14 small trucks

5 0

2 years ago

Read 2 more answers

What integer describes a decrease of seven? (A. 7 / B. -7 / C. 0 / D. 1/7)

Pie

Answer:

I thin B -7 cause its negative and negative goes down.

Step-by-step explanation:

3 0

2 years ago