Answer:
21 students pass
Step-by-step explanation:
Firstly, you can set up the problem into an equation where the variable X would equal the number of students passing. You put X over the total number of students in the class, turning it into a fraction, then set it equal to the fraction
(which is 75% represented as a fraction).

The fraction
can be simplified, because 75 and 100 are both multiples of 25, so after canceling out the 25s you would be left with
.

Next, you use the process of cross multiplication which is essentially just multiplying the denominators of both fractions (which would be 28 and 4 in this case) to each side of the equation.

The denominators cancel out leaving you with a simple equation to simplify.


Finally, divide both sides by four in order to isolate the variable.

X = 21.
Answer:
e) The mean of the sampling distribution of sample mean is always the same as that of X, the distribution from which the sample is taken.
Step-by-step explanation:
The central limit theorem states that
"Given a population with a finite mean μ and a finite non-zero variance σ2, the sampling distribution of the mean approaches a normal distribution with a mean of μ and a variance of σ2/N as N, the sample size, increases."
This means that as the sample size increases, the sample mean of the sampling distribution of means approaches the population mean. This does not state that the sample mean will always be the same as the population mean.
<span>1. Suppose that a family has an equally likely chance of having a cat or a dog. If they have two pets, they could have 1 dog and 1 cat, they could have 2 dogs, or they could have 2 cats.
What is the theoretical probability that the family has two dogs or two cats?
25% chance
</span><span>2. Describe how to use two coins to simulate which two pets the family has.
</span>
You could use the coins to simulate which pet the family has by flipping them and having head be dog and tails be cat (or vice-versa).
<span>3. Flip both coins 50 times and record your data in a table like the one below.
</span><span>Based on your data, what is the experimental probability that the family has two dogs or two cats?
</span>
Based on the results, I concluded that for Heads, Heads (which could be dogs or cats) there was a 24% chance and for Tails, Tails there was a 26% chance
<span>4. If the family has three pets, what is the theoretical probability that they have three dogs or three cats?
1/8 chance (accidentally messed up there) or 12.5%
</span><span>5. How could you change the simulation to generate data for three pets?
</span><span>
To flip 3 coins and add more spots on the chart.
I hope that this helps because it took a while to write out. If it does, please rate as Brainliest
</span>
Answer:do u mean divided 2?, then the answer is 325
Step-by-step explanation:650/2
Answer:c
Step-by-step explanation:
Given

and 
m of n is given by replacing x in
by x-3

can take any value of x except x=4 because at x=4 function is not defined
and
is also not defined for x=4 and can take any value of x