You measure 47 backpacks' weights, and find they have a mean weight of 79 ounces. Assume the population standard deviation is 13.1 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean backpack weight. Give your answer as a decimal, to two places

3.14

Step-by-step explanation:

Given that:

Mean weight (m) = 79 ounces

Population standard deviation (s) = 13.1

Sample size = 47

Maximal margin of error associated with 90% confidence interval.

The margin of error is given by:

Zcritical * (standard deviation / sqrt(sample size)

Z critical at 90% confidence interval = 1.645

Hence,

Zcritical * (standard deviation / sqrt(sample size)

1.645 * 13.1 / sqrt(47)

1.645 * (13.1 / 6.8556546)

1.645 * 1.9108313

Hence, the margin of error is :

3.1433174885

= 3.14

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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