There are 10 seniors in the class, from which 4 should be chosen by the teacher. The order of the chosen students does not matter. This means that we speak of combinations. THe equation for calculating the number of possible combinations is:
C=N!/R!(N-R), where N is the total number of objects and R is the number of objects we select from the N
In our case, N=10, R=4.
C= 10!/4!*6!=10*9*8*7*6!/6!*4*3*2*1=10*9*8*7/24=5040/24=210
There are 210 different ways for the teacher to choose 4 seniors in no particular order.

6 0

4 years ago

Select all of the following statments that are true

Nonamiya [84]

Answer:

A. -¾ + 0 = -¾

B. -¾ - ¾ = -(¾ + ¾)

C. ¾ - ¾ = ¾ + (-¾)

E. -¾ + ¾ = ¾ + (-¾)

F. -¾ + ¾ = 0

Step-by-step explanation:

Let's check each equation to determine whether they are true or false.

If what we have in the both sides are equal, then the equation is true, if they're not, them it is false.

✔️-¾ + 0 = -¾

Add everything on your left together

-¾ = -¾ (TRUE)

✔️-¾ - ¾ = -(¾ + ¾)

Add everything on both sides together respectively

(-3 - 3)/4 = -(3 + 3)/4

-6/4 = -6/4 (TRUE)

✔️¾ - ¾ = ¾ + (-¾)

0 = ¾ - ¾ (+ × - = -)

0 = 0 (TRUE)

✔️-¾ + ¾ = ¾ - (-¾)

0 = ¾ + ¾ (- × - = +)

0 = 6/4 (FALSE)

✔️-¾ + ¾ = ¾ + (-¾)

0 = ¾ - ¾ (+ × - = -)

0 = 0 (TRUE)

✔️-¾ + ¾ = 0

0 = 0 (TRUE)

6 0

3 years ago

A CBS News/New York Times opinion poll asked 1,190 adults whether they would prefer balancing the federal budget over cutting ta

nikklg [1K]

Using the normal distribution and the central limit theorem, it is found that there is a 0.0166 = 1.66% probability of a sample proportion of 0.59 or less.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sampling proportions of a proportion p in a sample of size n has mean $\mu = p$ and standard error $s = \sqrt{\frac{p(1 - p)}{n}}$