48% of the children in a summer camp are boys. If there are 1536 boys, how many children are there in the summer camp?

In Vancouver, British Columbia, the probability of rain during a winter day is 0.42, for a spring day is 0.23, for a summer day

nadezda [96]

Answer:

31.82% probability that this day would be a winter day

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening

In this question:

Event A: Rain

Event B: Winter day

Probability of rain:

0.42 of 0.25(winter), 0.23 of 0.25(spring), 0.16 of 0.25(summer) or 0.51 of 0.25(fall).

So

$P(A) = 0.42*0.25 + 0.23*0.25 + 0.16*0.25 + 0.51*0.25 = 0.33$

Intersection:

Rain on a winter day, which is 0.42 of 0.25. So

$P(A \cap B) = 0.42*0.25 = 0.105$

If you were told that on a particular day it was raining in Vancouver, what would be the probability that this day would be a winter day?

$P(B|A) = \frac{0.105}{0.33} = 0.3182$

31.82% probability that this day would be a winter day

6 0

2 years ago

Which of the following are true of F(x)?

aleksklad [387]

The answer is A the first choice

7 0

2 years ago

Read 2 more answers

Megan and Rosa are throwing sandbags during a workout. Megan's farthest throw measured 8.9 meters. Rosa's farthest throw measure

ankoles [38]

10.05-8.9=1.15. Rosa threw it 1.15 meters farther than Meghan

6 0

2 years ago

What is the best first step to take in solving the equation<br> x2+11x+24/ x2+3x =0

storchak [24]

x=root[3]{-1.500}=-1.14471

4 0

2 years ago

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

3 years ago