We would have the following sample space:
(1, 1), (1, 2), (1, 3), (1, 4)
(2, 1), (2, 2), (2, 3), (2, 4)
(3, 1), (3, 2), (3, 3), (3, 4)
(4, 1), (4, 2), (4, 3), (4, 4)
Those give us these sums:
2, 3, 4, 5
3, 4, 5, 6
4, 5, 6, 7
5, 6, 7, 8
P(sum of 2) = 1/16 =0.0625
P(sum of 3) = 2/16 = 0.125
P(sum of 4) = 3/16 = 0.1875
P(sum of 5) = 4/16 = 0.25
P(sum of 6) = 3/16 = 0.1875
P(sum of 7) = 2/16 = 0.125
P(sum of 8) = 1/16 = 0.0625
Answer:
Addition property of equality.
Hope this helps :)
I don’t kno how to estimate using benchmarks but I kno the exacts answer is 13/10 or 1.3
Answer:
c) A matched-pairs t-interval for a population mean difference
Step-by-step explanation:
Matched-Pairs t-Test. A matched-pairs t-test is used to test whether there is a significant mean difference between two sets of paired data.
Answer:
Machine's useful number of years = 9 years
Step-by-step explanation:
Using the straight line method, depreciation is calculated as the difference between the cost of the equipment minus the salvage value, all divided by the number of useful years.
Yearly Depreciation
= (Cost - Salvage value) ÷ (Number of useful years)
Yearly depreciation = P20,000
Cost = P200,000
Salvage Value = P20,000
Number of useful years = n
20000 = (200000 - 20000) ÷ n
20000 = (180000/n)
n = (180000/20000) = 9 years
Hope this Helps!!!
