Answer:
- 0.100
Step-by-step explanation:
Length of the ladder, H = 6 m
Distance at the bottom from the wall, B = 1.3 m
Let the distance of top of the ladder from the bottom at the wall is P
Thus,
from Pythagoras theorem,
B² + P² = H² .
or
B² + P² = 6² ..............(1) [Since length of the ladder remains constant]
at B = 1.3 m
1.3² + P² = 6²
or
P² = 36 - 1.69
or
P² = 34.31
or
P = 5.857
Now,
differentiating (1)
at t = 2 seconds
change in B = 0.3 × 2= 0.6 ft
Thus,
at 2 seconds
B = 1.3 + 0.6 = 1.9 m
therefore,
1.9² + P² = 6²
or
P = 5.69 m
on substituting the given values,
2(1.9)(0.3) + 2(5.69) × 
= 0
or
1.14 + 11.38 × = 0
or
11.38 × = - 1.14
or
= - 0.100
here, negative sign means that the velocity is in downward direction as upward is positive
Parallel has the same slope
2y + 8x = 18
Make it slope in form: y = mx + b
Divide the equation by 2
y + 4x = 9
Now subtract 4x
Y = -4x + 9
The slope is -4
Which equation also has same slope?
Solution: A
Answer:
196
Step-by-step explanation:
Given that the significance level of 2% = α
Because the the null hypothesis is true so it is denoted by α it also defined as the level of significance =1 - confidence level = 1 - 0.02 = 0.98.
So the number of the tests will incorrectly find significance:
=0.98×200=196
Answer:
There is a 24.3% probability that one of the calculators will be defective.
Step-by-step explanation:
For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The probability of a defective calculator is 10 percent.
This means that 
If 3 calculators are selected at random, what is the probability that one of the calculators will be defective
This is P(X = 1) when n = 3. So
There is a 24.3% probability that one of the calculators will be defective.
