Find the surface area of the triangular prism. The base of the

The zoo feeds the tortoise, rhinoceros, and zebra a total of 23 pounds of vegetables a day. How many pounds does the zoo feed th

vekshin1

If I understand your question correctly, you want to know how many pounds of vegetables the zoo feeds the animals a month if its a 30 day month?

In that case it would be 23 pounds times 30 days which is 690 pounds of vegetables. SO the zoo feeds the animals 690 pounds of vegetables to the animals each month.

Is that what your asking for?

8 0

2 years ago

Dakota’s parents are creating her practice ballet studio and want to install baseboards around the perimeter of the rectangular

kherson [118]

1. When you estimate, it is not an exact measurement. 3ft 8 in gets rounded to 4ft and 12 ft 3 in rounds to 12ft. now find the perimeter. P=2l+2w P= 2*12 +2*4 P=32feet

2. 3ft 8in = 3 8/12 or reduced to 3 2/3 12ft 3in = 12 3/12 or reduced to 12 1/4 The fractional part is referring to a fraction of a foot.

3. The perimeter of the room is P=2l+2w or P=2(12 1/4) + 2(3 2/3) p=24 1/2 + 7 1/3 P= 31 5/6 feet

4. The estimate and the actual are very close. They are 1/6 of a foot apart.

5a. Total baseboard 31 5/6ft - 2 1/4 ft = 29 7/12 feet needed.

5b. Take the total and divide it by 8ft = 29 7/12 divided by 8= 3.7 You are not buying a fraction of a board so you would need 4 boards.

4 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:

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