Answer:
B
Step-by-step explanation:
First calculate BD using sine ratio in Δ BCD and the exact value
sin60° =
, thus
sin60° =
=
=
=
( cross- multiply )
2BD = 12
( divide both sides by 2 )
BD = 6
-----------------------------------------------------------
Calculate AD using the tangent ratio in Δ ABD and the exact value
tan30° =
, thus
tan30° =
=
=
=
( cross- multiply )
AD = 6
( divide both sides by
)
AD = 6 → B
Answer:
(C)y=0
Step-by-step explanation:
An exponential function of the form
always has a horizontal asymptote at y = c.
Given our function 
Comparing with the form,
, we observe that c=0.
Therefore, the exponential function has an <u>asymptote at y=0.</u>
The correct option is C.
1. Take out the constants
-(2 x 3 x 4 x 2)xxyy^3
2. Simplify 2 x 3 x 4 x 2 to 48
-48xxyy^3
3. Use Product Rule: x^ax^b = x^a+b
-48x^1+1y^1+3
4. Simplify 1 + 1 to 2
-48x^2y^1+3
5. Simplify 1 + 3 to 4
-48x^2y^4
Answer:
0.35% of students from this school earn scores that satisfy the admission requirement.
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 zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
The combined SAT scores for the students at a local high school are normally distributed with a mean of 1479 and a standard deviation of 302.
This means that 
The local college includes a minimum score of 2294 in its admission requirements. What percentage of students from this school earn scores that satisfy the admission requirement?
The proportion is 1 subtracted by the pvalue of Z when X = 2294. So



has a pvalue of 0.9965
1 - 0.9965 = 0.0035
0.0035*100% = 0.35%
0.35% of students from this school earn scores that satisfy the admission requirement.