Answer: $72
Step-by-step explanation:
Simple interest (I) is calculated as Principal × Rate × Time.
From the question, Principal = $800
Rate =3% = 3/100= 0.03
Time =3 years
Slot in the values into the formula;
I = 800 x 0.03 x 3
I= $72
I hope this is clear
-- Look at the places where the line crosses the 'x' axis and the 'y' axis.
Between those two points, it goes forward 2 units, but only rises 1 unit.
So the slope of the line is 1/2 .
-- Look again at the point where the line crosses the y-axis.
At that point, y=1 .
That's the "y-intercept".
Now you have the slope and y-intercept of the line.
The equation of ANY straight line is
y = (slope)x + (y-intercept) .
You know what they both are, so you can easily write
y = 1/2 x + 1 .
Is that one of the choices ?
Yes it is.
Good enough for me.
That must be the correct one.
With no "blah blah blah" required.
Answer:
The answer is D. 5000
Step-by-step explanation:
(10)(10)(10)= 1000
1000(5)=5000
Answer:
-2.7 < y
Step-by-step explanation:
2.9 < 5.6+y
Subtract 5.6 from each side
2.9-5.6 < 5.6-5.6+y
-2.7 < y
Answer:
a)
b)
c)
With a frequency of 4
d)
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:
Replacing we got:
Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:
Part c
The mode is the most repeated value in the sample and for this case is:
With a frequency of 4
Part d
The midrange for this case is defined as:
Part e
For this case we can calculate the deviation given by:
And replacing we got:
And we can find the limits without any outliers using two deviations from the mean and we got:
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case