Answer: 1692
Step-by-step explanation:
Formula to find the sample size :

Given : Confidence level : 
⇒ significance level =
z-value for 90% confidence interval (using z-table)=
Prior estimate of the population proportion (p) of customers who keep up with regular vehicle maintenance is unknown.
Let we take p= 0.5
Margin of error : E= 2%=0.02
Now, the required sample size will be :

Simplify , we get

Hence, the required sample size = 1692
Answer:
-13x+10
Step-by-step explanation:
(4x+5)(x-2)
Multiply each term in the first parenthesis by each term in the second (foil)
Step 1: Expand it by writing out each multiplication. I added a picture showing which order to do it. (go in order of green, red, blue, yellow.) (you can remember this as first, outside, inside, last.)
When you expand it'll look like: 4x⋅x+4x⋅-2-5x-5⋅-2
Step 2: Calculate product
4x⋅x+4x⋅-2-5x-5⋅-2 (for the 4x⋅x it would be written like
)
+4x⋅-2-5x-5⋅-2
+4x⋅-2-5x-5⋅-2 (4x⋅-2 becomes -8x) (multiply 4x times -2)
-8x-5x-5⋅-2
-8x-5x-5⋅-2 (-5⋅-2 becomes +10) (multiply -5 times -2)
-8x-5x+10
Step 3: collect like terms
-8x-5x+10 (-8x-5x becomes -13x) (-8x times -5x)
-13x+10 is the most simplified so it should be your final answer
In rectangle ABCD
AB * AC = area
In another rectangle WXYZ
WX * WY = area
The sides must be equal or have same multiples and should be divisible by each other
True - since the outlier is way off it will cause your average to mess up which is why we test multiple times to make sure we are more accurate with our numbers
Correct Ans:Option A. 0.0100
Solution:We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.