Hello!
The number of tickets is dependent on age, so age is the independent variable.
Think of it as a y = mx equation. In this equation, x is the input, and y changes depending on the value of x inputted.
In this situation, y is the number of tickets, as y changes depending on the age x inputted. x is age, the independent variable.
Answer:
Step-by-step explanation:
The wording on this is not the best. It sounds like the 1 zero has even multiplicity (that's because of where the modifier is). On top of that it has an odd power. You could try this. y =x*(x^2+1)^2
The problem is not with the power. It gives x^5. The problem is with the multiplicity of the one place where it crosses. (X^2 + 1) does factor, but it gives a complex root. I'm not sure that's allowed. However, it is the best I can do.
D)30 us 6 times as many as 5 and 5 times as many as 6.
Answer:
= 2sinA.cosA+ sinA / 1+cos2A+ cosA
= sinA(2cosA + 1) / 2cos^2A + cosA
= sinA(2cosA + 1) / cosA( 2cosA + 1)
= sinA / cosA
= tan A
= R.H.S
Answer:
The probability will be 0.3085 or 0
Step-by-step explanation:
Given:
True mean=12.5
Sample mean =12.6
Standard deviation=0.2
Samples=100
To Find:
Probability that exceeds 12.6 ounces.
Solution:
Calculate the Z-score for given means and standard deviation.
So
Z-score= (true mean -sample mean)/standard deviation.
Z-score=(12.5 -12.6)/0.2
=-0.1/0.2
=-0.5
Now Using Z-table
P(X≥-0.5)=p(Z≥-0.5)=0.3085
Hence Probability that sample mean weight exceeds will be 0.3085
OR
By using Normal distribution with sampling ,it will be as follows
Z=(X-u)/[Standard deviation/Sqrt(No of samples)]
Z=(12.6-12.5)/(0.2/Sqrt(100)
Z=0.1/0.2/10
Z=5
So P(X≥12.6 )=P(Z≥5)=1
Pr(Z≥5)=1-1=0.
(Refer the attachment )
Hence Probability of getting ounces greater than 12.6 is '0'.
The sampling is of 0.02 size hence graphically it looks likely.
as shown in attachment.