Answer:
On 5th day and 13th day Attendance was equal for both plays and it was
231 for each .
Step-by-step explanation:
In order to find same attendance ,we need plug both equation equal
-x^2+26x+126 = 8x+191
-x^2+26x-8x+126-191 = 0
-x^2 +18x-65 =0
multiplying both sides by -1 ,we get
x^2 -18x+ 65 =0
(x-5) (x-13) =0
x =5 and x =13
On 5th day and 13th day attendance was same.
And attendance can be obtained by plugging x =5 in either of equation
y = 8(5)+191
= 40 +191
= 231
A) 
B)In 200 times he can hit 59 times !
<u>Step-by-step explanation:</u>
Here we have , A baseball player got a hit 19 times in his last 64 times at bat. We need to find the following :
a. What is the experimental probability that the player gets a hit in an at bat?
According to question ,
Favorable outcomes = 19
Total outcomes = 64
Probability = (Favorable outcomes)/(Total outcomes) i.e.
⇒ 
⇒ 
b. If the player comes up to bat 200 times in a season, about how many hits is he likely to get?
According to question , In 64 times he hit 19 times . In 1 time there's probability to hit 0.297 times! So ,In 200 times he can hit :
⇒ 
⇒ Hit = 59.36
Therefore , In 200 times he can hit 59 times !
Cp=7512.75×[100÷(100+77÷4)]
=7512.75×(100÷119.25)
=6300
c.p=6300