What is 2 + 4+6x9+12?

Four less than the product of one and number x

MaRussiya [10]

1x-4 I hope this helps.

6 0

3 years ago

Help please!!! I would appreciate it!

sertanlavr [38]

A.) 5 Units
B.) 5 Units
C.) 5 Units
D.) 5 Units

7 0

3 years ago

Read 2 more answers

Exhibit: Department Store.

aalyn [17]

Answer:

Step-by-step explanation:

Considering the central limit theorem, the distribution is normal since the number of samples is large.

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Confidence interval = mean ± z × σ/√n

Where

σ = population standard Deviation

σ/√n = sample standard deviation

Confidence interval = x ± z × σ/√n

1) x = $75

σ = $24

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.96 = 0.04

α/2 = 0.04/2 = 0.02

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.02 = 0.98

The z score corresponding to the area on the z table is 2.05. Thus, confidence level of 96% is 2.05

Confidence interval = 75 ± 2.05 × 24/√64

= 75 ± 2.05 × 3

= 75 ± 6.15

The lower end of the confidence interval is

75 - 6.15 = 68.85

The upper end of the confidence interval is

75 + 6.15 = 81.15

2) n = 400

x = $75

σ = $24

z = 2.05

Confidence interval = 75 ± 2.05 × 24/√400

= 75 ± 2.05 × 1.2

= 75 ± 2.46

The lower bound of the confidence interval is

75 - 2.46 = 72.54

The upper bound of the confidence interval is

75 + 2.46 = 77.46

3) n = 400

x = $200

σ = $80

The z score corresponding to the confidence level of 94% is 1.88

z = 1.88

Confidence interval = 200 ± 1.88 × 80/√400

= 200 ± 1.88 × 4

= 200 ± 7.52

Margin of error = 7.52

7 0

3 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

!!!!!Help!!!!

Alja [10]

The correct answer is y=-13/10x+13 as it matches the graph

Hope this helped :)

7 0

3 years ago