It is quadratic, because the highest degree for x is 2, x². Expand the expression and you'll see: f(x)=3x²-3x-3x-7
simplify: f(x)=3x²-6x-7
the quadratic term is 3x², the linear term is -6x, the constant is -7
Considering that the subjects are chosen without replacement, they are not independent, and the probability cannot be found using the binomial distribution.
The binomial distribution and the hypergeometric distribution are quite similar, as:
- They find the probability of exactly x successes on n repeated trials.
- For each trial, there are only two possible outcomes.
- The difference is that the binomial distribution is for independent trials, that is, in each trial, the probability of success is the same, while the hypergeometric distribution is for dependent trials.
- If the sample is without replacement, the trials are not independent, thus the hypergeometric distribution is used, not the binomial.
A similar problem is given at brainly.com/question/21772486
positive = greater than 0.
Therefore we have the inequality:
|add 7 to both sides
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
D=number of dimes
q=number of quarters
let's count everything in cents
dimes are worth 10 cents
quarters are worth 25 cents
10d+25q=700
divide both sides by 5
2d+5q=140
he has 7 more quarters than dimes
q=7+d
subsitute 7+d for q in other equation
2d+5q=140
2d+5(7+d)=140
2d+35+5d=140
7d+35=140
minus 35 both sides
7d=105
divide both sides by 7
d=15
sub back
q=7+d
q=7+15
q=22
22 quarters and 15 dimes
