Step-by-step explanation:
If the figure is triangle then,
Let PR be x
So,
PQ+QR+PR= 180 (Sum of all the angles of the triangle)
Therefore,
6x+25+16-3x+x=180
4x+36=180
4x= 180/36
x= 144/4
X= 36
NOW,
PQ= 6x+25
= 6(36)+25
= 216+25 = 241= PQ.
QR = 16-3x
= 16-3(36)
= 16-108
= -92
PR = 36
hope it helps
Answer: Hello there!
initially Maria has 6 red marbles, 6 blue marbles, and 4 green marbles.
You know that in the first pick, Maria gets a green marble ( this means that now there are 3 green marbles left)
Now we want to know the probability that she gets a red marble on the second pick.
Now there are 6 + 6 + 3 = 15 marbles in the bag, and 6 of them are red ones. Then the probability of getting one red is the number of red ones divided by the total amount ( which is the ratio of red marbles to total marbles)
this is 6/15.
now you can see that there is no 6/15 in the options, so we need to see which one is equivalent.
if we took the option C, 2/5, and multiply both denominator and numerator by 3, we get (3*2)/(3*5) = 6/15, then 2/5 and 6/15 are equivalents.
Then the correct answer is C.
Step-by-step explanation:
Hope this is the right one.
Problem 8p^2 - 30p + c
<em>Step One</em>
Take 1/2 of - 30
1/2 * -30 = - 15
<em>Step 2
</em>Square -15
(-15)^2 = 225
c = 225
Problem Nine
a = 1
b = 4
c = -15


x = [-4 +/- sqrt(76)] / 2
x = [-4 +/- 2*sqrt19]/2
x = [-4/2 +/- 2/2 sqrt[19]
x = - 2 +/- sqrt(19)
x1 = - 2 + sqrt(19)
x2 = -2 - sqrt(19)
These two can be broken down more by finding the square root. I will leave them the way they are. It's just a calculator question if you want it to go into decimal form.
Problem Tena = 1
b = 4
c = -32
The discriminate is sqrt(b^2 - 4ac)
D = sqrt(b^2 - 4ac)
D = sqrt(4^2 - 4(1)(-32)
D = sqrt(16 - - 128)
D = sqrt(16 + 128)
D = sqrt(144)
D = +/- 12
Since D can equal + or minus 12 there must be 2 possible (and different) roots. As a matter of fact, this quadratic can be factored.
(x + 8)(x - 4) = y
But that' s not what you were asked for.
The discriminate is > 0 so the roots are going to be real.
<em>
Answer; The discriminate is > 0 so there will be 2 real different roots.</em>
Answer:
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
Step-by-step explanation:
denoting A= a piece is defective , Bi = a piece is defective from the i-th supplier and Ci= choosing a piece from the the i-th supplier
then
P(A)= ∑ P(Bi)*P(C) with i from 1 to 3
P(A)= ∑ 5/100 * 24/100 + 10/100 * 36/100 + 6/100 * 40/100 = 9/125
from the theorem of Bayes
P(Cz/A)= P(Cz∩A)/P(A)
where
P(Cz/A) = probability of choosing a piece from Z , given that a defective part was obtained
P(Cz∩A)= probability of choosing a piece from Z that is defective = P(Bz) = 6/100
therefore
P(Cz/A)= P(Cz∩A)/P(A) = P(Bz)/P(A)= 6/100/(9/125) = 5/6 (83.33%)
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
The interquartile range is 4