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3 years ago

What is 2/5 +1/3= show your work

Mathematics

2 answers:

Scrat [10]3 years ago

6 0

Must find a common denominator

2/5 + 1/3

= 6 + 5/15
= 11/15

natita [175]3 years ago

3 0

The first thing you want to do is get a common denominator. This means the number at the bottom of both fractions is the same.

In this problem it would be fifteen
2/5 x 3/3 = 6/15
1/3 x 5/5 = 5/15

When adding fractions you only add the numerators or the top numbers together. When adding fractions the denominator is NEVER ADDED.
6/15 + 5/15 = 11/15

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Wayne Gretsky scored a Poisson mean six number of points per game. sixty percent of these were goals and forty percent were assi

BartSMP [9]

Answer:

a) The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

b) 0.05 = 5% probability that he has four goals and two assists in one game

Step-by-step explanation:

In hockey, a point is counted for each goal or assist of the player.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval. The standard deviation is the square root of the mean.

(a) Find the mean and standard deviation for the total revenue he earns per game.

60% of six are goals, which means that 60% of the time he earned 3K.

40% of six are goals, which means that 40% of the time he earned 1K.

The mean is:

$\mu = 6*0.6*3 + 6*0.4*1 = 13.2$

The standard deviation is:

$\sigma = \sqrt{\mu} = \sqrt{13.2} = 3.63$

The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

(b) What is the probability that he has four goals and two assists in one game

Goals and assists are independent of each other, which means that we find the probability P(A) of scoring four goals, the probability P(B) of getting two assists, and multiply them.

Probability of four goals:

60% of 6 are goals, which means that:

$\mu = 6*0.6 = 3.6$

The probability of scoring four goals is:

$P(A) = P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.19122$

Probability of two assists:

40% of 2 are assists, which means that:

$\mu = 6*0.4 = 2.4$

The probability of getting two assists is:

$P(B) = P(X = 2) = \frac{e^{-2.4}*(2.4)^{2}}{(2)!} = 0.26127$

Probability of four goals and two assists:

$P(A \cap B) = P(A)*P(B) = 0.19122*0.26127 = 0.05$

0.05 = 5% probability that he has four goals and two assists in one game

5 0

3 years ago

For what values of a are the following expressions true?/a+5/=-5-a

katovenus [111]

Explanation:

The expression is given below as

$|a+5|=-5-a$

Concept:

We will apply the bsolute rule below

$\begin{gathered} if|u|=a,a>0 \\ then,u=a,u=-a \end{gathered}$

By applying the concept, we will have

$\begin{gathered} \lvert a+5\rvert=-5-a \\ a+5=-5-a,a+5=5+a \\ a+a=-5-5,a-a=5-5 \\ 2a=-10,0=0 \\ \frac{2a}{2}=\frac{-10}{2},0=0 \\ a=-5,0=0 \end{gathered}$

Hence,

The final answer is