Answer:
JL = 68
Step-by-step explanation:
JM = ML
Hence, 3x + 4 = 5x - 16
Subtract 3x from both sides
4 = 2x - 16
Add 16 to both sides
20 = 2x
Divide both sides by 2
10 = x
Now we want to find JL
Note that JL = JM + ML
JM = 3x + 4 and x = 10
Substitute 10 for x
3(10) + 4
Multiply
30 + 4
Add
JM = 34
Because JM = ML
We simply multiply 34 times 2 to find the length of JL
34 * 2 = 68
JL = 68
Answer:
7
Step-by-step explanation:
1 glass of juice cost $3
1 glass of limonade costs $2
So the proportion between them (Orange) / (juice) = 3/2 or 3¹/₂
2x - 3y = 6
-3y = -2x + 6
y = 2/3x - 2....the slope here is 2/3. A parallel line will have the same slope
y = mx + b
slope(m) = 2/3
(9,-3)...x = 9 and y = -3
now we sub and find b, the y int
-3 = 2/3(9) + b
-3 = 6 + b
-3 - 6 = b
-9 = b
so ur parallel equation is : y = 2/3x - 9 <== or 2x -3y = 27
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!