All categories

3 years ago

Find the volume of the cone. Leave your answers in therms of pi.

Mathematics

1 answer:

SOVA2 [1]3 years ago

5 0

Hey there! :)

The formula for finding volume of a cone = (π · r² · h) ÷ 3

r = 3
3² = 3 × 3 = 9

h = 7

7 × 3 × π = 21π

Your answer is C. 21π³

Hope this helps :)

You might be interested in

Factorise the following expressions 5x^2-25xy

Natali [406]

Answer: 5x ( x - 5y )

Step-by-step explanation:

3 0

3 years ago

One 3.8 lb bag of fertilizer covers 1000 square feet each bag cost $9.95 the square feet of the yard is 1307.25 ft squared how m

zaharov [31]

Answer:

Exactly 1.30725 bags are needed to cover the garden.

Step-by-step explanation:

We have a yard with a surface of 1307.25 square feet.

We know that one bag of fertilizer covers 1000 square feet.

We want to know how many bags are needed to cover the garden. We can calculate the number of bags as:

$B=S\cdot b=1307.25\;\text{sq.ft.}\cdot \dfrac{1\,\text{bag}}{1000\,\text{sq.ft.}}=1.30725\,\text{bags}$

B: number of bags needed for the yard.

S: surface of the yard.

b: number of bags per square foot.

4 0

3 years ago

Solve each equation 1. 3/4 f + 5 = -5 2.-1/5 b - 2/5 = -2 Please show work

anzhelika [568]

Sole for f by simplifying both sides of the equation then isolating the variable so that means f=-40 over 3 for number one

for the second equation you do the same thing you did for number one and the answer will be
b=8

4 0

3 years ago

105.25 x 70.75 NO LINKS!!!!!

Anarel [89]

Answer:

7446.4375

Step-by-step explanation:

105.25

x 70.75

-------------

52625

First, we will want to start with the hundredths place. multiply 5×5. We get 25. Put the 5 down and carry the 2 above 2 in 105.25. Next, we multiply 2 in 105.25 with 5 in 70.75. 2×5 is 10, then we add the carried 2, and 10+2 is 12. We put the 2 down below, and carry the 1 above 5 in 105.25. Next, we multiply the 5 in 105.25 times 5. 5×5 is 25. We then need to add the carried 1 to 25 to get 26. We put the 6 down and put the 2 over 0 in 105.25. Next, we multiply 0 in 105.25 with 5 in 70.75. 0×5 is 0, plus our carried 2 from before, so 2. We don't need to carry anything since it's only the ones digit, so we put the 0 down. Next, we multiply the 1 in 105.25 with 5 again in 70.75. 1×5=5. Since we have no carries we can put the 5 down without adding anything. If this seemed confusing, let me simplify it for you. You basically start with the bottom number's "digit". For us, that's 70.75, and 5 would be the "digit." (The last digit from the bottom number.) We then multiply this "digit" by every number from the top number's digits. For example, we'd multiply 70.75's 5, with 105.25's 5 first, then continue on going left from there. I hope this helped! If it just confused you more please let me know. (Also remember to put a 0 when you start multiplying with your second digit on the bottom number)

3 0

3 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago