According to a candy company, packages of a certain candy contain 1818% orange candies. find the approximate probability that the random sample of 100100 candies will contain 2121% or more orange candies.

0.2033

Step-by-step explanation:

Given that in a candy company orange candies are 18% of total.

Proportion p is N (p,

i.e. p is Normal with mean = 0.18 and std dev = 0.036

Hence probabilities that 100 candles will contain 2121% or more orange candies.

=P(

Z score for 0.21 = \frac{0.21-0.18}{0.036}

=0.8333

P(Z>=0.83) = 0.5-0.2967=0.2033

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116