Step 1
<u>Find the value of TS</u>
we know that
if PQ is parallel to RS. then triangles TRS and TPQ are similar
so

solve for TS

we have


substitute

Step 2
<u>Find the value of SQ</u>
we know that

we have


substitute

therefore
<u>the answer is</u>
the value of SQ is 
Answer:
1. x2 - 9 > 0
x^2-3^2>0
(x+3)(x-3)>0
(x+3)>0 and (x-3)>0
x>-3 and x>3
2. x2 - 8x + 12 > 0
x^2 - 8x +12>0
x^2 -2x -6x +12 >0 (-8x is replaced by (-2x) + (-6x) )
x(x-2) -6(x-2) >0
(x-6)(x-2)>0
(x-6)>0 and (x-2)>0
x>6 and x>2
3. -x2 - 12x - 32 > 0
-x^2 -12x -32 >0
x^2 +12x +32 <0
x^2 +4x +8x +32<0
x(x+4) +8(x+4)<0
(x+8)(x+4)<0
(x+8)<0 and (x+4)<0
x<-8 and x<-4
4. x2 + 3x - 20 >= 3x + 5
x^2 +3x -20 >= 3x +5
x^2 +3x -20 -3x >= 3x +5 -3x
x^2 -20 >= 5
x^2 -20 +20 >= 5 +20
x^2 >=25
x^2-25 >=0
(x-5)(x+5)>=0
(x-5)>=0 and (x+5)>=0
x>=5 and x>=-5
The answer to this question is D ( $0.18 ) per ounce:-)
Answer:
The probability that at most 38 of the 60 relays are from supplier A is P(X≤38)=0.3409.
Step-by-step explanation:
We can model this question with a binomial distribution random variable.
The sample size is n=60.
The probability that the relay come from supplier A is p=2/3 for any relay.
If we use a normal aproximation, we have the mean and standard deviation:

The probability that at most 38 of the 60 relays are from supplier A is P(X≤38)=0.3409:

So what do you need us to do for you?