Question

1. the frequency table shows the results of a survey of a certain forest. find the probability that a randomly chosen tree will be a california redwood tree taller than 300 ft. round to the nearest thousandth.

trees in forest

height of 300 ft	over	under	totals
giant sequoia	35	270	305
totals	60	450	510

a. partial b. 0.082 c. 0.417 d. 0.041

1. the table shows the results of a survey of students in two math classes. the students were asked whether they watched more than one hour of tv the night before.

find p(more than 1 hr. of tv | 6th period class). round to the nearest thousandth.
hint: make sure you get the totals first!

tv survey results

	yes	no
6th period class	11	8

a. 0.550 b. 0.727 c. 0.579 d. 0.421

1. in how many different orders can you line up 8 cards on a table?

a. 8 b. 1 c. 1,680 d. 40,320

1. in how many different orders can 12 basketball players be listed in a program?

a. 665,280 b. 1 c. 479,001,600 d. 12

1. partial which...

Explanation:

Question 9

Step1: Identify relevant values

We need the number of California Redwood trees taller than 300 ft (which is 25) and the total number of trees (which is 610).

Step2: Calculate probability

Probability is the number of favorable outcomes (California Redwood > 300 ft) divided by the total number of outcomes (total trees). So, \( P=\frac{25}{610}\approx0.041 \)

Step1: Find totals for 6th period class and total students

For 6th period class: Yes = 11, No = 8, so total for 6th period is \( 11 + 8=19 \). Total students: 3rd period (10 + 10 = 20) + 6th period (19) = 39? Wait, no, let's check the table again. Wait, the table is: 3rd period: Yes=10, No=10; 6th period: Yes=11, No=8. So total "Yes" is \( 10 + 11 = 21 \), total "No" is \( 10+8 = 18 \), total students \( 21 + 18 = 39 \)? Wait, no, the question is "P(More than 1 hr of TV | 6th period class)". So it's a conditional probability. The number of 6th period students who watched more than 1 hr (Yes) is 11, and the total number of 6th period students is \( 11 + 8 = 19 \). So probability is \( \frac{11}{19}\approx0.579 \)

Step2: Verify calculation

\( \frac{11}{19}\approx0.5789\approx0.579 \)

Step1: Recall permutation of n items

The number of ways to line up \( n \) cards is \( n! \) (n factorial), where \( n! = n\times(n - 1)\times\cdots\times1 \). Here, \( n = 8 \), so \( 8! = 8\times7\times6\times5\times4\times3\times2\times1 \)

Step2: Calculate 8!

\( 8! = 40320 \)? Wait, no, wait 5! is 120, 6! is 720, 7! is 5040, 8! is 40320? Wait, but the options have D. 40,320? Wait, no, the options are A. 8, B. 1, C. 1,680, D. 40,320. Wait, 8! is 40320, which is option D? Wait, no, wait maybe I misread. Wait the question is "In how many different orders can you line up 8 cards on a table?" So it's 8 permutations of 8, which is \( _8P_8=\frac{8!}{(8 - 8)!}=8! = 40320 \)

Answer:

D. 0.041

Question 10