Answer:
c- shifted 3 units right and 4 units up
Step-by-step explanation:
In this problem, we have a quadrilateral named as ABCD. Recall that a quadrilateral is a two-dimensional shape having four sides. So, we need to identify what transformation has been performed to get A'B'C'D', which is the same quadrilateral shifted certain units right and up. So take one point, say, B, so how do we need to do to obtain point B'? well, we need to move that point 3 units right and 4 units up, but how can we know this? just count the number of squares you need to move from B to B' horizontally and vertically, which is in fact 3 units right and 4 units up.
It would have taken you 5:30 for each mile. 55/10 is 5.5 thats 5 and a half, half a minute is 30 seconds. so 5 minutes and 30 seconds
For this case, the first thing we are going to do is write the generic equation of motion for the vertical axis.
We have then:

Where,
- <em>g: acceleration of gravity
</em>
- <em>vo: initial speed
</em>
- <em>h0: initial height
</em>
For the first body:

For the second body:

By the time both bodies have the same height we have:


Rewriting we have:



Clearing time:

Answer:
it takes 18.31s for the two window washers to reach the same height
Answer:
0.108
Step-by-step explanation:
Using the poisson probability process :
Where :
P(x =x) = (e^-λ * λ^x) ÷ x!
Given that :
Each batch of bread = 3 loaves
Each loaf = 15 slices
Total slice per batch = 15 * 3 = 45 slices
Number of raising added = 100
Average number of raisin per slice, λ = 100/45 = 20/9
Hence,
Probability that a randomly chosen slice has no raising :
P(x = 0) = (e^-λ * λ^x) ÷ x!
P(x = 0) = (e^-(100/45) * (100/45)^0) ÷ 0!
P(x = 0) = (0.1083680 * 1) / 1
P(x = 0) = 0.108
Answer:
The 95% of confidence intervals
(2.84 ,2.99)
Step-by-step explanation:
A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16
given small sample size n =20
sample mean x⁻ =2.92
sample standard deviation 'S' =0.16
level of significance ∝ = 0.95
The 95% of confidence intervals
the degrees of freedom γ=n-1 =20-1=19
t-table 2.093


(2.92-0.0748,2.92+0.0748)
(2.84 ,2.99)
Therefore the 95% of confidence intervals
(2.84 ,2.99)