Answer:
b. exam completion time is negatively skewed
Step-by-step explanation:
A data distribution is said to be negatively skewed when <em>median</em> value of the distribution is higher than the <em>average</em> value of the distribution.
In this example
- average mid-term completion time is 40 minutes
- median mid-term completion time is 55 minutes.
thus median > mean, so the mid-term completion time is negatively skewed. Negatively skewed distributions are also called left-skewed.
Answer:
Option A
Step-by-step explanation:
Number of components assembled by the new employee per day,
N(t) = 
Number of components assembled by the experienced employee per day,
E(t) = 
Difference in number of components assembled per day by experienced ane new employee
D(t)= E(t) - N(t)
D(t) = 
= 
= 
= 
= 
Therefore, Option A will be the answer.
You are lazy... do it on your own.
Answer:
The coefficient 6 in 6x means 6(x) or 6 multiplied by x 6 times.
Step-by-step explanation:
X could be anything, it is only holding the place of a number. The coefficient 6 before the x is a number being multiplied by the unknown variable, x. For example if x = 2 the expression would turn into 6(2) which equals 12.
Hope this helps! If you have any other questions please don't hesitate to ask me or your teacher so you can be sure to master the subject. :)
Answer:
1. 1343 years
2. 9 hours
3. 39 years
Step-by-step explanation:
1. Given, half-life of carbon = 5730 years.
∴ λ = 0.693/half-life of carbon = 0.693/5730 = 0.000121
If N₀ = 100 then N = 85
Formula:- N = N₀*e^(-λt)
∴ 85 = 100 * e^(-0.000121t)
∴㏑(-0.85)=-0.000121t
∴ t = 1343 years
2. Given half-life of aspirin = 12 hours
λ = 0.693/12 = 0.5775
Also N₀ = 100 then 70 will disintegrate and N = 30 will remain disintegrated.
∴ 70 = 100 *e^(-0.05775t)
0.70 = e^(-0.05775t)
㏑(0.70) = -0.05775t
∴ t = 9 hours
3. The population of the birds as as A=A₀*e^(kt)
Given that the population of birds fell from 1400 from 1000, We are asked how much time it will take for the population to drop below 100, let that be x years.
The population is 1400 when f = 0, And it is 1000 when f = 5
We can write the following equation :
1400 = 1000e^(5t).
∴1400/1000 = e^(5k)
∴ k = ㏑(1.4)/5
We need to find x such that 1400/100 = e^(xk)
14 = e^(xk)
∴ x = 39 years