Volume= a³
a=edge
2197 = a³
=> a=13 m
Surface area=6a²=6*13²=6*169=1014 m²
 $\frac{1014}{40}$ = 25,35 ≈ 26
=> 26 cans needed

8 0

3 years ago

HELP PLZ!!!!- In 2012, on average, about 9.46 × 10^-1 pound of potatoes was produced for every 2.3 × 10^-5 power acre harvested.

sashaice [31]

Answer: The answer is 41.13 pounds (approx.).

Step-by-step explanation: Given that the average production of potatoes in 2012 was 0.946 pounds for every 0.000023 power acre harvested. We need to calculate the pounds of potatoes produced on average for each acre harvested.

For that, we need to divide the pounds of potatoes by number of acres harvested.

In 0.000023 acre harvested, ponds of potatoes = 0.946 ponds.

Therefore, in 1 acre harvested, pounds of potatoes is given by

$p=\dfrac{0.946}{0.000023}=41.13~\textup{(approx.)}$

Thus, the answer is 41.13 pounds (approx.).

3 0

3 years ago

Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its competitor's. A sample of 40 bab

Readme [11.4K]

Answer:

Mean weight gained of two goods is not significantly different under 0.05 or 0.01 significance level, but it is under 0.10 significance level.

Step-by-step explanation:

We need to calculate the z-statistic of the differences of sample means and compare if it is significant under a significance level.

Z-score can be calculated using the formula:

z= $\frac{X-Y}{\sqrt{\frac{s(x)}{N(x)}+\frac{s(y)}{N(y)}}}$ where

X is the mean weight gain for in the first three months after birth for babies using the Gibbs products.
Y is the mean weight gain for in the first three months after birth for babies using the competitor products
s(x) is the population standard deviation of the sample for Gibbs brand
s(y) is the population standard deviation of the sample for competitor brand
N(x) is the sample size for babies used Gibbs product
N(y) is the sample size for babies used competitor product.

putting the numbers in the formula:

z= $\frac{7.6-8.1}{\sqrt{\frac{2.3}{40}+\frac{2.9}{55}}}$ ≈ -1.51

and z-table gives that P(z<-1.51) = 0.0655

To conclude if the competitor good is significantly better, we need to choose a significance level and compare it to 0.0655.

For example, the difference in mean weight gained of two goods is not significant under 0.05 or 0.01 significance since 0.0655 is bigger than these values. But we can conclude that under 0.10 significance, competitor brand mean weight gain is significantly more than the Gibbs brand mean weight gain.

5 0

4 years ago