Calculate ten point two minus three point nine

Solve for x.

antoniya [11.8K]

Answer:

3x/5 - 367/100

Step-by-step explanation:

4 0

2 years ago

Can someone help me

Mashcka [7]

Answer:

whois someone? OH ! Right, That dude named Some Won . Sorry.. i'll let him answer.

Step-by-step explanation:

8 0

3 years ago

Need help with another question please

hoa [83]

The answer for this is m
=
5.436564
v
2
?

3 0

2 years ago

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

2 years ago

A researcher records the odometer reading and age of used Hondas. What kind of correlation is likely to be obtained for these tw

tresset_1 [31]

Using the concept of correlation, it is found that a strong positive correlation is expected between these two variables.

When two variables are direct proportional, that is, both increase together, there is a strong positive correlation.

In this problem, the older the car, the more it should have traveled, that is, the higher the read on the odometer, hence, there is a direct proportional relationship, which means that the variables have a strong positive correlation.

A similar problem is given at brainly.com/question/15468813

5 0

2 years ago