"a day care program has an average daily expense of $75.00. the standard deviation is $5.00. the owner takes a sample of 64 bills. what is the probability the mean of his sample will be between $70.00 and $80.00? "

p=1

Step-by-step explanation:

Given that a day care program has an average daily expense of $75.00. The standard deviation is $5.00.

Sample size = 64

Std error of sample =

Thus sample mean will follow normal with

(75, 0.625)

Required probability

=

It is almost a certain event with p =1

their you go i had it wrong so thought i give someone the right answer

Step-by-step explanation:

1.0

34

1.0

34

68

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -