Let William be x years old.
Thomas = (4 + x) years
In five years,
William = (x + 5) years
Thomas = 4 + x + 5 = (9 + x) years
Given, in five years sum of their ages is 24.
x + 5 + 9 + x = 24
2x + 14 = 24
2x = 24 - 14
2x = 10
x =
x = 5
Hence, William is 5 years old and Thomas is 9 years old.
Answer <u>(assuming it is allowed to be in point-slope format)</u>:
Step-by-step explanation:
1) First, determine the slope. We know it has to be perpendicular to the given equation, 
. That equation is already in slope-intercept form, or y = mx + b format, in which m represents the slope. Since 
is in place of the m in the equation, that must be the slope of the given line.
Slopes that are perpendicular are opposite reciprocals of each other (they have different signs, and the denominators and numerators switch places). Thus, the slope of the new line must be 
.
2) Now, use the point-slope formula, 
to write the new equation with the given information. Substitute 
, 
, and 
for real values.
The represents the slope, so substitute in its place. The and represent the x and y values of a point the line intersects. Since the point crosses (1,4), substitute 1 for and 4 for . This gives the following equation and answer:
Answer:
3/2
Step-by-step explanation:
The equation 3x-2y=4 can be written in the form of y=mx+c where m is the slope and c is the y intercept.
Therefore
2y=3x-4 and dividing both sides by 2 to have y at the LHS
y=3/2x-2
Therefore, 3/2 represents the gradient/slope
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
Answer:Part A: 13
Step-by-step explanation: look at the graph
