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3 years ago

How many liters of water can a hemisphere with a diameter of 30 cm hold?

Mathematics

1 answer:

ZanzabumX [31]3 years ago

3 0

Answer:

The figure 25.4 cm is suggestive of metric equivalent of 10″ and coupled with the fact the thickness of the bowl is not considered implies some estimate will do. Just take diameter to be 1/4 m and get volume of (22/42) x8 = 4.25 liters correct to 5%.

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Solve 4x - 3 = 18. Round to the nearest thousandth.

lorasvet [3.4K]

Answer:

$x=5.250$

Step-by-step explanation:

We have been given an equation $4x-3=18$ . We are asked to solve our given equation.

First of all, we will add 3 to both sides of our equation.

$4x-3+3=18+3$

$4x=21$

Now, we will divide both sides of our equation by 4.

$\frac{4x}{4}=\frac{21}{4}$

$x=5.25$

Upon rounding our answer to nearest thousandth (3 places after decimal) we will get,

$x=5.250$

Therefore, the solution for our given equation is $x=5.250$ .

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Answer:

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2 years ago

Organizar de menor a mayor las fracciones 1/2 1/4 ,3/4,,1

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A student takes a driving test until it is passed.

Leya [2.2K]

Answer:

The probability that the test is taken an even number of times is 0.30.

Step-by-step explanation:

The probability that a student passes the driving test at any attempt is,

<em>p</em> = 4/7.

The event of a student passing in any attempt is independent of each other.

The probability that the test is taken an even number of times is:

P (even number of tests) = P (Passing in the 2nd attempt)

+ P (Passing in the 4th attempt)

+ P (Passing in the 6th attempt) ...

If a student passed in the 2nd attempt it implies that he failed in the first.

Then, P (Passing in the 2nd attempt) = $(\frac{3}{7}) \times (\frac{4}{7})$

Similarly, P (Passing in the 4th attempt) = $(\frac{3}{7})^{3} \times (\frac{4}{7})$ , since he failed in the first 3 attempts.

And so on.

Compute the probability of an even number of tests as follows:

P (even number of tests) = $(\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...$

The result follows a Geometric progression for infinite values.

The sum of infinite GP is:

$S=\frac{a}{1-r^{2}}$

The probability is:

P (even number of tests) = $(\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...$

$=\frac{(\frac{3}{7})(\frac{4}{7} ) }{1-(\frac{3}{7})^{2}}\\=\frac{12}{49}\times\frac{49}{40}\\ =\frac{12}{40}\\ =0.30$

Thus, the probability that the test is taken an even number of times is 0.30.

7 0

3 years ago