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Nikitich [7]
3 years ago
8

5,000 is 1/10 of what number

Mathematics
2 answers:
Andrej [43]3 years ago
6 0
5000 is 1/10 of what number....

5000 = 1/10x
5000 / (1/10) = x
5000 * 10/1 = x
50,000 = x <=== ur number
Sergeeva-Olga [200]3 years ago
6 0
5,000 is 1/10 of 50,000. An easy way to figure it out is 5,000x10=50,000 to check that you put 50,000/10=5,000.
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What is the volume of a hemisphere with a radius of 9.8in, rounded to the nearest tenth of a cubic inch?
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<u>Given</u>:

Given that the radius of the hemisphere is 9.8 inches.

We need to determine the volume of the hemisphere.

<u>Volume of the hemisphere:</u>

The volume of the hemisphere can be determined using the formula,

V=\frac{2}{3} \pi r^3

where r is the radius of the hemisphere.

Substituting r = 9.8 in the above formula, we get;

V=\frac{2}{3} (3.14)(9.8)^3

Simplifying, we get;

V=\frac{2}{3} (3.14)(941.192)

V=\frac{5910.69}{3}

V=1970.23

Rounding off to the nearest tenth, we get;

V=1970.2 \ in^3

Thus, the volume of the hemisphere is 1970.2 cubic inches.

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It is impossible to slice a triangular pyramid with a plane and get a quadrilateral cross section.
Ksivusya [100]

Answer:

false

Step-by-step explanation:

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cam hits the bullseye in 8 darts out of 15 throws. what is the experimental probability that came next throws will hit the bulls
Zigmanuir [339]

Answer:

A darts player practices throwing a dart at the bull’s eye on a dart board. Her probability of hitting the bull’s eye for each throw is 0.2.

(a) Find the probability that she is successful for the first time on the third throw:

The number F of unsuccessful throws till the first bull’s eye follows a geometric

distribution with probability of success q = 0.2 and probability of failure p = 0.8.

If the first bull’s eye is on the third throw, there must be two failures:

P(F = 2) = p

2

q = (0.8)2

(0.2) = 0.128.

(b) Find the probability that she will have at least three failures before her first

success.

We want the probability of F ≥ 3. This can be found in two ways:

P(F ≥ 3) = P(F = 3) + P(F = 4) + P(F = 5) + P(F = 6) + . . .

= p

3

q + p

4

q + p

5

q + p

6

q + . . . (geometric series with ratio p)

=

p

3

q

1 − p

=

(0.8)3

(0.2)

1 − 0.8

= (0.8)3 = 0.512.

Alternatively,

P(F ≥ 3) = 1 − (P(F = 0) + P(F = 1) + P(F = 2))

= 1 − (q + pq + p

2

q)

= 1 − (0.2)(1 + 0.8 + (0.8)2

)

= 1 − 0.488 = 0.512.

(c) How many throws on average will fail before she hits bull’s eye?

Since p = 0.8 and q = 0.2, the expected number of failures before the first success

is

E[F] = p

q

=

0.8

0.2

= 4.

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who knowsAnswer:

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