Answer:
Solution given:
perpendicular distance =height [p]=40m
base or horizontal distance [b]=198m
Hypotenuse or zip line [h]=?
Now
By using Pythagoras law
h²=p²+b²
h²=40²+198²
h=√40804
h=202m
<u>So</u><u> </u><u>the</u><u> </u><u>Zip</u><u> </u><u>line</u><u> </u><u>is</u><u> </u><u>2</u><u>0</u><u>2</u><u>m</u><u> </u><u>long</u><u>.</u>
Answer:
BBAB
Step-by-step explanation:
DU DU DU DU DU DU DU DU DU
Answer:
6:10pm
Step-by-step explanation:
At 6:00 the hands are 180° apart.
The hour hand moves 1/12 of circle per hour = 30° per hour = 0.5° per minute.
The minute hand moves 1/60 of the circle per minute = 6° per minute
Angle between the hands decreases by 6-0.5 = 5.5° per minute.
180 - 5.5t = 125
t = 125/5.5 = 10 minutes
The hands are 125° apart at t minutes after 6:00, or 6:10.
Answer:
a) The probability that a box containing 3 defectives will be shipped is
b) The probability that a box containing only 1 defective will be sent back for screening is
Step-by-step explanation:
Hi
a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so
The second step is to count the number of total possible random sets of taking a sample size of 4 items over 20 items of the box (discounting the 3 defectives) as the possible ways to succeed, so
Finally we need to compute , therefore the probability that a box containing 3 defectives will be shipped is
a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so
The second step is to count the number of total possible random sets of taking a sample size of 4 items over 22 items of the box (discounting the defective 1) as the possible ways to succeed, so
Then we need to compute , therefore the probability that a box containing 1 defective will be shipped is
Finally the probability that a box containing only 1 defective will be sent back for screening will be
Answer: three sets of value show 3 consecutive increases and they could be the intensities during fourth, fifth, and six visits:
- 66%, 69%, 72%;
- 63%, 65%, 67%, and
- 67%, 72%, 77%
Explanation:
1) The program recommends a constant intensity for 3 visits, which is what the table shows:
Day Intensisty
1 63%
2 equal ⇒ 63%
3 equal ⇒ 63%
2) Hence, you have to determine the valid sets that meet the recommendation for the fourth, fifth, and six visits, which are the next three.
2) For the next three visits, the program recommensd increasing intensities.
There are three options that show 3 consecutive increases; they are:
- 66%, 69%, 72%;
- 63%, 65%, 67%, and
- 67%, 72%, 77%
Therefore, those are the choices that apply.