Answer:
it the one that say he didn times 2 an eigt togeter incorrect
Step-by-step explanation:
Add: 2/9 + 3/5 = 2 · 5/9 · 5 + 3 · 9/5 ·9 = 10/45 + 27/45 = 10 + 27/45 = 37/45
Answer:
There is a 1.1% chance that a randomly chosen junior will have a score above 95.
Step-by-step explanation:
Calculate the z-score of 95 given a normal distribution with mean 79 and standard deviation of 7. The z-score is a probability (area under the distribution curve) of a value on a normalized random variable that has a distribution with mean 0 and standard deviation of 1. To get such a transformation, you need to subtract the mean and divide by the standard deviation, like this:
Based on this z-value, use z-score tables to look up the area under the curve. This will be the probability that randomly chosen values are lower (or higher, depending which table you are using) than this z-score. Be careful to apply the correct table. I found the following probability that a random z value is higher that the z-score 2.29: 0.011, or 1.1%. This means that there is a 1.1% chance that a randomly chosen junior will have a score above 95.
40 when you join and the rest is paid monthly.Write an equation to represent the situation that can help members find how much they pay per month
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.
