F is number of frogs sold
p is number of Popsicles sold
6f+1.5p=94.50
f+p=36
Rearrange
p=36-f
Substitute
6f+1.5(36-f)=94.5
Reduce
6f+54-1.5f=94.5
4.5f=40.5
F=9
Substitute
9+p=36
P=27
George sold 27 Popsicles and 9 Frogs
There are two parts in the given collection. One contains quarters and other contains dimes.
Let's compare the whole coin collection to the quarters.
One way to represent the ratio is 15 : 7.
The other way is 
Answer:
a) Statement is true
b) Statement incorrect. Company goal is to attend customer at 190 second at the most. So test should be one tail test (right)
c) We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error
Step-by-step explanation:
a) as the significance level is = 0,02 that means α = 0,02 (the chance of error type I ) and the β (chance of type error II is
1 - 0.02 = 0,98
b) The company establishe in 190 seconds time for customer at its ticket counter (if this time is smaller is excellent ) company is concerned about bigger time because that could be an issue for customers. Therefore the test should be a one test-tail to the right
c) test statistic
Hypothesis test should be:
null hypothesis H₀ = 190
alternative hypothesis H₀ > 190
t(s) = ( μ - μ₀ ) / s/√n ⇒ t(s) = ( 202 - 190 )/(28/√100 )
t(s) = 12*10/28
t(s) = 4.286
That value is far away of any of the values found for 99 degree of fredom and between α ( 0,025 and 0,01 ). We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error
If we look table t-student we will find that for 99 degree of freedom and α = 0.02.
Answer:
a) ⅖+⅓ = 6/15 + 5/15 = 11/15
b) 5/9 - 1/2 = 10/18 - 9/18 = 1/18
