Catering company provides packages for weddings and for showers. The cost per person for small groups is pproximately Normally distributed for both weddings and showers. The mean cost for weddings is $82.30 with a
andard deviation of $18.20, while the mean cost for showers is $65 with a standard deviation of $17.73. If 9
eddings and 6 showers are randomly selected, what is the probability the mean cost of the weddings is more than
e mean cost of the showers?

2

step-by-step explanation:

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

the number of the people going.

step-by-step explanation:

the cost to buy one movie ticket is $7. if the total cost for the movie is a function of how many people go, then the input is : the number of the people going.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15