Answer: the equation is slope-intercept form y = -2/3x
(2x - 5) (5x - 10)
Use FOIL
10x² - 20x - 25x + 50
10x² - 45x + 50
5( 2x² - 9x + 10)
2x -5
x -2
5(2x - 5)(x -2)
make each equal to 0
2x - 5 = 0
x - 2 = 0
2x - 5 = 0 x - 2 = 0
2x - 5 (+5) = 0 x - 2 (+2) = 0 (+2)
2x = 5 x = 2
x = 5/2
x = 5/2 , 2
hope this helps
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.
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01

X[bar] ± 
174.5 ± 
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!
Answer:
65/100
Step-by-step explanation:
Since it is a whole percentage it would be over 100. Since its 65%, it would be 65/100