$141, this is because 235 divide by 10 is 23.5, multiply 23.5 by 6 and your answer is $141
Answer:
Step-by-step explanation:
The identities you need here are:
and 
You also need to know that
x = rcosθ and
y = rsinθ
to get this done.
We have
r = 6 sin θ
Let's first multiply both sides by r (you'll always begin these this way; you'll see why in a second):
r² = 6r sin θ
Now let's replace r² with what it's equal to:
x² + y² = 6r sin θ
Now let's replace r sin θ with what it's equal to:
x² + y² = 6y
That looks like the beginnings of a circle. Let's get everything on one side because I have a feeling we will be completing the square on this:

Complete the square on the y-terms by taking half its linear term, squaring it and adding it to both sides.
The y linear term is 6. Half of 6 is 3, and 3 squared is 9, so we add 9 in on both sides:

In the process of completing the square, we created within that set of parenthesis a perfect square binomial:

And there's your circle! Third choice down is the one you want.
Fun, huh?
2x+4y=16
x-7=-4y
4y=16-2x
<span>4y=16-2(-4y+7)
4y=16+8y-14
-4y=2
-4y/-4=2/-4
<u>y=-0,5</u>
2x=16-4(-0,5)
2x=16+2
2x/2=18/2
<u>x=9
</u>
C</span>
Answer:
38 = 1 x 38 or 2 x 19. Factors of 38: 1, 2, 19, 38. Prime factorization: 38 = 2 x 19
Answer:
IQ scores of at least 130.81 are identified with the upper 2%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 100 and a standard deviation of 15.
This means that 
What IQ score is identified with the upper 2%?
IQ scores of at least the 100 - 2 = 98th percentile, which is X when Z has a p-value of 0.98, so X when Z = 2.054.




IQ scores of at least 130.81 are identified with the upper 2%.