Answer:
a

b

c
Option A is correct
Step-by-step explanation:
From the question we are told that
The sample size is n = 15
The probability of success is 
The number of success we are considering is r = 10
Now the probability of failure is mathematically evaluated as

substituting value


Now using the binomial distribution to find the probability of exactly 10 successes we have that
![P(X = r ) = [\left n } \atop {r}} \right. ] * p^r * q^{n- r}](https://tex.z-dn.net/?f=P%28X%20%3D%20%20r%20%29%20%3D%20%20%5B%5Cleft%20n%20%7D%20%5Catop%20%7Br%7D%7D%20%5Cright.%20%5D%20%2A%20p%5Er%20%2A%20%20q%5E%7Bn-%20r%7D)
substituting values
![P(X = 10 ) = [\left 15 } \atop {10}} \right. ] * p^{10}* q^{15- 10}](https://tex.z-dn.net/?f=P%28X%20%3D%20%2010%20%29%20%3D%20%20%5B%5Cleft%2015%20%7D%20%5Catop%20%7B10%7D%7D%20%5Cright.%20%5D%20%2A%20p%5E%7B10%7D%2A%20%20q%5E%7B15-%2010%7D)
Where
mean 15 combination 10 which is evaluated with a calculator to obtain
![[\left 15 } \atop {10}} \right. ] = 3003](https://tex.z-dn.net/?f=%5B%5Cleft%2015%20%7D%20%5Catop%20%7B10%7D%7D%20%5Cright.%20%5D%20%20%3D%203003)
So


Now using the normal distribution to approximate the probability of exactly 10 successes, we have that

Applying continuity correction

substituting values


Standardizing

The where
is the mean which is mathematically represented as

substituting values


The standard deviation is evaluated as

substituting values


Thus



From the normal distribution table we obtain the
as

And the 

There value can also be obtained from a probability of z calculator at (Calculator dot net website)
So


Looking at the calculated values for question a and b we see that the values are fairly different.
Answer:
The jar of chermical contains 66g of Zinc.
Step-by-step explanation:
The ratio of Zinc to Copper is 3 : 13
A jar of the chemical contains 286g of copper.
Let the mass of the Zinc be xg.
= 
x =
= 66g
So the jar of chemical contains 66g of Zinc.
Answer:
subtract 4 from both sides
Answer:
ef + eg
Step-by-step explanation:
Distributive Property: a(b +c) = (a*b) + (a*c)
e(f +g) = e*f + e*g
= ef + eg