Answer:
The random variable (number of toppings ordered on a large pizza) has a mean of 1.14 and a standard deviation of 1.04.
Step-by-step explanation:
<em>The question is incomplete:</em>
<em>The probability distribution is:</em>
<em>x P(x)
</em>
<em>
0 0.30
</em>
<em>1 0.40
</em>
<em>2 0.20
</em>
<em>3 0.06
</em>
<em>4 0.04</em>
The mean can be calculated as:

(pi is the probability of each class, Xi is the number of topping in each class)
The standard deviation is calculated as:

Answer:

Step-by-step explanation:
Sample size 52 card
Pay for J or Q 
Pay for King or Ace 
Pay for others 
Therefore
Probability of drawing J or Q

Probability or drawing King or Ace

Probability or drawing Other cards

Therefore
Expected Gain is mathematically given as



Answer: About 83915 grams
.............................
The answer is D. The number of AP tests increases as GPA increases.