Combination of normal and binomial distributions.
Calculate probability of opening below 100m.
mean=175 m
standard deviation = 35 m
z=(100-175)/35=-15/7
p=P(Z<z)=0.01606229 from normal probability tables.
Calculate probability of at least one parachute opens below 100m
p=0.01606229n=5
x=0
P(X>0)
=1-P(x=0)
=1-C(5,0)p^x(1-p)^5
=1-1*0.01606229^0*(1-0.01606229)^5
=1-0.9222 (approximately)
=0.0778
Answer:
Probability that at least one of five cargoes will be damaged is 0.0778
E is the answer (sorry if this is wrong)
Answer:
B
Step-by-step explanation:
Using the rule of logarithms
x = n ⇔ x = 
Given
= 35, then
35 = a, that is
ln35 = a → B
Y = -x + 2 is your answer.
This is because the slope is -1 and the y-intercept is 2.
Answer:
10800 cm³ oil can the small container carry.
Option (A) is correct .
Step-by-step explanation:
The scale factor is defined by

As given
The scale factor of a large oil container to a small oil container is 0.075.
The large oil container can carry 144,000 cm³ of oil.
Scale factor = 0.075
Larger volume = 144000 cm³
Put in the above

Small volume = 144000 × 0.075
Small volume = 10800 cm³
Therefore 10800 cm³ oil can the small container carry.
Option (A) is correct .