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Brrunno [24]
3 years ago
7

What is 20% of 100 in whole money

Mathematics
2 answers:
Eduardwww [97]3 years ago
6 0
20\% \ of \ 100 = \\  \\ \frac{20}{\not100} \cdot\not100  = \\ \\ =\boxed{20}
Kryger [21]3 years ago
3 0
%=percent
percent means parts out of 100
20%=20/100
'of' means multiply

20% of 100  units means
20/100 times 100 units=2000/100 units=20 units
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A track star runs two races on a certain day. The probability thathe wins the first race is 0.7, the probability that he wins th
NARA [144]

Answer:

a) 80% probability that he wins at least one race.

b) 30% probability that he wins exactly one race.

c) 20% probability that he wins neither race.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that he wins the first race.

B is the probability that he wins the second race.

C is the probability that he does not win any of these races.

We have that:

A = a + (A \cap B)

In which a is the probability that he wins the first race but not the second and A \cap B is the probability that he wins both these races.

By the same logic, we have that:

B = b + (A \cap B)

The probability that he wins both races is 0.5.

This means that A \cap B = 0.5

The probability that he wins the second race is 0.6

This means that B = 0.6

B = b + (A \cap B)

0.6 = b + 0.5

b = 0.1

The probability that he wins the first race is 0.7.

This means that A = 0.6

A = a + (A \cap B)

0.7 = a + 0.5

a = 0.2

A) he wins at least one race.

This is

P = a + b + (A \cap B) = 0.2 + 0.1 + 0.5 = 0.8

There is an 80% probability that he wins at least one race.

B) he wins exactly one race.

This is

P = a + b = 0.2 + 0.1 = 0.3

There is a 30% probability that he wins exactly one race.

C) he wins neither race

Either he wins at least one race, or he wins neither. The sum of these probabilities is 100%.

From a), we have that there is an 80% probability that he wins at least one race.

So there is a 100-80 = 20% probability that he wins neither race.

6 0
3 years ago
Help please i don’t know how to do it
Norma-Jean [14]

Answer:

Step-by-step explanation:

\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\

From the initial condition,

\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C

So we have that \ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1

4 0
3 years ago
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