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Soloha48 [4]
3 years ago
6

The Barrel in Problems 4 is 5 feet long and the radius of its base is 1.5 feet. ​

Mathematics
1 answer:
Vsevolod [243]3 years ago
5 0
The answer that you are look big for would be 39.829 feet.
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Which is greater?<br><br> A. 3 1/2<br> or<br> B. 3 1/8
OleMash [197]

Answer:

3 1/2 = 7/2

3 1/8 = 25/8

7/2 x 4 = 28/8

7/2 (3 1/2) < 25/8 (3 1/8)

Therefore, 3 1/8 is greater than 3 1/2

Step-by-step explanation:

8 0
3 years ago
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Use the table below to determine the approximate height of the flag pole.
Artyom0805 [142]

Answer:

B 17.5 ft

Step-by-step explanation:

tangent = opposite / adjacent

tan 35° = h / 25

multiply both sides by 25

h = 25 (tan 35°)

tan 35° = 0.70 (from your calculator)

h = 25 (0.70)

h = 17.51 ft.

3 0
3 years ago
What is the difference of income tax, pay roll tax,sales tax, property tax
gayaneshka [121]

Answer:

Income tax refers to money the company owes based on its earnings. Sales tax refers to money the company collects from customers and sends to the state tax collector. Payroll taxes refer to money the company owes based on the wages it pays its employees.

Step-by-step explanation:

8 0
3 years ago
Y is inversely proportional to x.<br> When y=50, x=4 <br> Work out y when x=8
notka56 [123]

Answer:

100

Step-by-step explanation:

5 0
3 years ago
A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If
Blababa [14]

Answer: Our required probability is \dfrac{1}{2}

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = \dfrac{1}{3}

So, Probability of getting head from fair coin = \dfrac{1}{2}

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}

Hence, our required probability is \dfrac{1}{2}

No, the answer is not \dfrac{1}{3}

5 0
3 years ago
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