Jerry is paid $6 per hour. He worked 1 hour and 10 minutes.
So, we'll need to find out how much he is paid per minute. We'll divide $6 by 60 minutes; which will give us $0.10 per minute.
For the 10 minutes extra he worked, we'll multiply by $0.10.
10 x 0.10 = 1
This shows that he earned an extra $1 for the additional 10 minutes he worked.
He earned $6 for the full hour plus $1 for the additional 10.
6 + 1 = 7
Therefore, Jerry earned $7.00 for his time worked on that day.
Answer:
This is simple! (Kind of)
Step-by-step explanation:
First, notice how HJ is tangent. HG is a radius intersecting HJ at H.
This means, (According to some theorem that I forgot the name of) that GHJ is a right angle.
Thus, we can use the 180* in a triangle theorem.
So, let's solve!
So, there you go! Nice and simple!
Hope this helps!
Stay Safe!
Answer:
s = 2t
Step-by-step explanation:
At the end of a snow storm, Audrey saw there was a lot of snow on her front lawn. The temperature increased and the snow began to melt at a steady rate. There was a depth of 10 inches of snow on the lawn when the storm ended and then it started melting at a rate of 2 inches per hour. Write an equation for S, in terms of t, representing the depth of snow on Audreys lawn, in inches, t hours after the snow stopped falling.
2 inches = 1 hour
10 inches = x hour
2x = 10
x = 10/2
x = 5
Creating an equation
S ∝ t
S = kt
k = s/t
k = 2/1
k = 2
The equation for S, in terms of t, representing the depth of snow on Audreys lawn, in inches, t hours after the snow stopped falling is
S = 2t
Answer:
AB ,AE
Step-by-step explanation:
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.
