Mathematics College

leon wrote an expression that is equivalent to (30+6)÷12 witch expression could be the one leon wrote

Answers

Answer 1

ANSWER

(3 · 3 · 2 · 2) ÷ (3 · 2 · 2)

EXPLANATION

The given expression is also equivalent to 36 ÷ 12 - because 30 + 6 = 36.

In the equivalent expression we have:

[tex]\begin{gathered} 3\cdot3\cdot2\cdot2=36 \\ 3\cdot2\cdot2=12 \end{gathered}[/tex]

Therefore it's 36 ÷ 12 too

Related Questions

In the figure below, m2 = 49. Find mx 1.

Answers

By definition, a Right angle is an angle that measures 90 degrees.

Complementary angles are those angles that add up to 90 degrees.

For this case, you can identify that the angle 1 and the angle 2 are Complementary angles, because when you add them, you get 90 degrees (a Right angle).

Knowing the above, you can set up the following equation:

[tex]m\angle1+m\angle2=90\degree[/tex]

Since you know that:

[tex]m\angle2=49\degree[/tex]

You can substitute this value into the equation and the solve for the angle 1 in order to find its measure. You get that this is:

[tex]\begin{gathered} m\angle1+49\degree=90\degree \\ m\angle1=90\degree-49\degree \\ m\angle1=41\degree \end{gathered}[/tex]

The answer is:

[tex]m\angle1=41\degree[/tex]

The average score for games played in the NFL is 22 and the standard deviation is 9.3 points. 41 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution.

a. What is the distribution of ¯x x¯
? ¯xx¯ ~ N( , )
b. What is the distribution of ∑x ? ∑x ~ N ( , )
c. P( ¯x > 19.8214) =
d. Find the 60th percentile for the mean score for this sample size.
e. P(20.6214 < x¯< 23.2262) =
f. Q1 for the x¯distribution =
g. P( ∑x > 829.0774) =

For part c) and e), Is the assumption of normal necessary? NoYes

Answers

Using the normal distribution and the central limit theorem, it is found that:

a) The distribution is: x¯ ~ N(22, 1.45).

b) The distribution is: ∑x ~ N(902, 59.55).

c) P( ¯x > 19.8214) = 0.9332 = 93.32%.

d) The 60th percentile for the mean score for this sample size is of 22.37 points a game.

e) P(20.6214 < x¯< 23.2262) = 0.6312 = 63.12%.

f) Q1 for the x¯distribution = 21 points a game.

g) P( ∑x > 829.0774) = 0.8888 = 88.88%.

Assumption of normality is not necessary, as the sample sizes are greater than 30.

Normal Probability Distribution

The z-score of a measure X of a variable that has mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by the rule presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The z-score measures how many standard deviations the measure X is above or below the mean of the distribution, depending if the z-score is positive or negative.From the z-score table, the p-value associated with the z-score is found, and it represents the percentile of the measure X in the distribution.By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].Also by the Central Limit Theorem, for the sum of n instances of a variable, the mean is of [tex]\n\mu[/tex] and the standard deviation is of [tex]\sigma\sqrt{n}[/tex].Finally, by the Central Limit Theorem, assumption of normality is only necessary when the sample size is less than 30.

For a single game, the mean and the standard deviation of the number of points scored are given as follows:

[tex]\mu = 22, \sigma = 9.3[/tex]

For the average of 41 games, the standard error is:

[tex]s = \frac{9.3}{\sqrt{41}} = 1.45[/tex]

Hence the distribution is: x¯ ~ N(22, 1.45).

For the sum of the 41 games, the mean and the standard error are given as follows:

41 x 22 = 902.[tex]s = 9.3\sqrt{41} = 59.55[/tex].

Hence the distribution is: ∑x ~ N(902, 59.55).

In item c, the probability is one subtracted by the p-value of Z when X = 19.8214, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem: