Answer:
23.9894949495cm is the depth of the container.
Step-by-step explanation:
Cylindrical container:
height: 28
radius: 9 (or 18/2)
pi * 9² = 254.4690049408
254.4690049408 * 28 = 7,125.1321383424 cm²
cylidrical container area: 7,125.1321383424cm²
Container:
length: 27cm
width: 11cm
27 * 11 = 297
7,125.1321383424 / 297 = 23.9894949495cm
I assume that the container with rectangular base is precisely full when tou pour the water into it.
And that your question was how deep the container was.
To solve problem 19, we must remember the order of operations. PEMDAS tells us that we should simplify numbers in parentheses first, exponents next, multiplication and division after that, and finally addition and subtraction. Using this knowledge, we can begin to simplify the problem by working out the innermost set of parentheses:
36 / [10 - (3-1)²]
36 / [10 - (2)²]
Next, we should still simplify what is inside the parentheses but continue to solve the exponents (the next letter in PEMDAS).
36/ (10-4)
After that, we should compute the subtraction that is inside the parentheses.
36/6
Finally, we can solve using division.
6
Now, we can move onto problem 20:
1/4(16d - 24)
To solve this problem, we need to use the distributive property, which allows us to distribute the coefficient of 1/4 through the parentheses by multiplying each term by 1/4.
1/4 (16d-24)
1/4(16d) - 1/4(24)
Next, we can simplify further by using multiplication.
4d - 6
Therefore, your answer to problem 19 is 6 and the answer to problem 20 is 4d -6.
Hope this helps!
16/25 = 0.64 = 64%
Another way to do it is to multiply top and bottom by 4
16/25 = (16*4)/(25*4) = 64/100 = 64%
Either way, the answer is 64%
Answer:
3865.74 J/s
Step-by-step explanation:
mass of ice, m = 1 ton = 1000 kg
time , t = 24 hours
latent heat of fusion of ice, L = 334000 J/kg
Heat required to melt, H = m x L
where, m is the mass of ice and L be the latent heat of fusion
So, H = 1000 x 334000 = 334 xx 10^6 J
Rate of heat transfer = heat / time = 
Rate of heat transfer = 3865.74 J/s
thus, the rate of heat transfer is 3865.74 J/s.
Answer:
b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.
Step-by-step explanation:
Given that in a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.9 hours.
(Right tailed test at 5% level)
Mean difference = 0.2
Std error = 
Z statistic = 1.0540
p value = 0.145941
since p >alpha we do not reject H0.
b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.
