Answer is B, the graph increases by about 0.28 per year
Let the number of comic books be X
if we add 14 to X = 24
so we will find the value of X to get half initial number of comic books William had.
14+X=24
X=24-14
X=10
so the total innitial number of comic books William had is 10+10=20 books.
Answer:
The answer is 3rd point
Step-by-step explanation:
By looking at the table :
10-10.99,
$15.95 (1 dot)
20-29.99,
$21.52, $25.55, $27.78, $29.90 (4 dots)
30-39.99,
$33.98, $36.77 (2 dots)
40-49.99,
$43.44 (1 dot)
50-59.99,
$51.25 (1 dot)
60-69.99,
$60.54, $64.56, $67.89 (3 dots)
Then you can determine number of dots in each range.
Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:
And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
The probability that she picks two white socks is 0.0357
<h3>How to determine the probability?</h3>
The given parameters are:
Socks = 8
Black = 4
White = 2
Brown = 1
Red = 1
The probability of picking a white sock at first is:
P(White) = 2/8
Now, there are 7 socks left
The probability of picking a white sock next is
P(White) = 1/7
The required probability is:
P = 2/8 * 1/7
Evaluate
P = 0.0357
Hence, the probability that she picks two white socks is 0.0357
Read more about probability at:
brainly.com/question/251701
#SPJ1
