Which inequality is true when x=2.5

Answer?plz. Need help

Brrunno [24]

Answer:

C.

Step-by-step explanation:

The GCF of 28 and 12 is 4

Soo... Distribute it to find the factors. 28/4=7 and 12/4 is 3.

4(7+3)

6 0

2 years ago

4.

avanturin [10]

Answer:

A.) x / 3 = 12 ----> x = 12(3) ----><em> </em><em>x = 36</em>

B.) 2x + 3 = 20 ----> 2x = 17 ----> x = 17/2

C.) 4/3x = 10/3 ----> x = 10/3 x 3/4 ----> x = 30/12 = 5/2

D.) -4x = -24 ----> x = -24/-4 ----> x = 6

E.) 2(x-4) = 10 ----> 2x - 8 = 10 ----> 2x = 18 ----> x = 9

F.) -0.5x + 1.1 = -2.9 ----> -0.5x = -4 ---> x = -4 / -0.5 ----> x = 8

4 0

3 years ago

A normally distributed data set has a mean of 0 and a standard deviation of 2. Which of the following is closest to the percent

frutty [35]

A normally distributed data set has a mean of 0 and a standard deviation of 2. The closest to the percent of values between -4.0 and 2.0 would be 84%.

<h3>What is the empirical rule?</h3>

According to the empirical rule, also known as the 68-95-99.7 rule, the percentage of values that lie within an interval with 68%, 95%, and 99.7% of the values lies within one, two, or three standard deviations of the mean of the distribution.

$P(\mu - \sigma < X < \mu + \sigma) \approx 68\%\\P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 95\%\\P(\mu - 3\sigma < X < \mu + 3\sigma) \approx 99.7\%$

A normally distributed data set has a mean of 0 and a standard deviation of 2.

$Z=(x-\mu)/\sigma$

$P(x=-0.4)\\\\z=(-0.4-0)/2\\\\= -0.2\\\\P(x=2)\\z=(2-0)/2\\\\=1$

$P(-0.4 < x < 2)=p(-0.2 < z < 1)=p(-0.2 < z < 0)+p(0 < z < 1)$ ……….(by symmetry)

=.49865+.3413

.83995…….(by (http://83995…….by) table value)

=.8400 × 100

=84%

Learn more about the empirical rule here:

brainly.com/question/13676793

#SPJ2

4 0

1 year 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:

$\mu = 6*0.6 = 3.6$