We learned in exercise 4.18 that about 90% of american adults had chickenpox before adulthood. we now consider a random sample of 120 american adults. (a) how many people in this sample would you expect to have had chickenpox in their childhood? and with what standard deviation? (b) would you be surprised if there were 105 people who have had chickenpox in their childhood? (c) what is the probability that 105 or fewer people in this sample have had chickenpox in their childhood? how does this probability relate to your answer to part (b)?

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !

answer is 8x+24

area of vanessa's patio = length × width

given, length = (3x² + 5x + 10)

width = (x² - 3x - 1)

area of vanessa's patio = (3x² + 5x + 10)(x² - 3x - 1)

= 3x²(x² - 3x - 1) + 5x(x² - 3x - 1) + 10(x² - 3x - 1)

= 3x⁴ - 9x³ - 3x² + 5x³ - 15x² - 5x + 10x² - 30x - 10

= 3x⁴ - 4x³ - 8x² - 35x - 10

hence, expression of vanessa's patio = (3x⁴ - 4x³ - 8x² - 35x - 10)

a

step-by-step explanation: