Answer:
D
Step-by-step explanation:
Pretty sure it's correct
Answer:
- 12x +9y ≥ 510
- y ≤ 2x
- y ≥ 25
Step-by-step explanation:
Phoebe earns 12x for her yard weeding, and Richie earns 9y for his dog walking. They want the sum of these earnings to be at least 510, so the first constraint inequality is ...
12x +9y ≥ 510
__
Richie plans to walk no more than 2x dogs, since y is the number of dogs he walks, the second constraint is ...
y ≤ 2x
__
Richie will walk at least 25 dogs, so ...
y ≥ 25
Answer:
The correct option is (c).
Step-by-step explanation:
The complete question is:
The data for the student enrollment at a college in Southern California is:
Traditional Accelerated Total
Math-pathway Math-pathway
Female 1244 116 1360
Male 1054 54 1108
Total 2298 170 2468
We want to determine if the probability that a student enrolled in an accelerated math pathway is independent of whether the student is female. Which of the following pairs of probabilities is not a useful comparison?
a. 1360/2468 and 116/170
b. 170/2468 and 116/1360
c. 1360/2468 and 170/2468
Solution:
If two events <em>A</em> and <em>B</em> are independent then:

In this case we need to determine whether a student enrolled in an accelerated math pathway is independent of the student being a female.
Consider the following probabilities:

If the two events are independent then:
P (F|A) = P(F)
&
P (A|F) = P (A)
But what would not be a valid comparison is:
P (A) = P(F)
Thus, the correct option is (c).
1)
LHS = cot(a/2) - tan(a/2)
= (1 - tan^2(a/2))/tan(a/2)
= (2-sec^2(a/2))/tan(a/2)
= 2cot(a/2) - cosec(a/2)sec(a/2)
= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))
= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)
= 2[(1+cos(a) -1)/sin(a)]
=2cot a
= RHS
2.
LHS = cot(b/2) + tan(b/2)
= [1 + tan^2(b/2)]/tan(b/2)
= sec^2(b/2)/tan(b/2)
= 1/sin(b/2)cos(b/2)
using product to sums
= 2/sin(b)
= 2cosec(b)
= RHS
Answer:
Figure RST is congruent is R´S´T´
Step-by-step explanation:
Given that R´S´T is a reflection of RST, we know that is a rigid transformation, so it is congruent (the same size & shape).