How heavy a load (pounds) is needed to pull apart pieces of douglas fir 4 inches long and 1.5 inches square? here are data from students doing a laboratory exercise:
33,190 31,860 32,590 26,520 33,280
32,320 33,020 32,030 30,460 32,700
23,040 30,930 32,720 33,650 32,340
24,050 30,170 31,300 28,730 31,920
suppose that the strength of pieces of wood like these follows a normal distribution with standard deviation 3000 pounds. give a 90% confidence interval for the mean load required to pull the wood apart.

answer: the answer is 23

step-by-step explanation:

answer: sensy

step-by-step explanation:

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.

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14

step-by-step explanation:

14