Y=0.2x+0.3=0.5
y=0.05x+0.4=0.09
all you have to do is basically just add the numbers when you add the numbers that would be your fraction you have to add the zeros and which will make the number
=50-(50*30%)
=50-(50*0.30)
=50-15
=$35
the new cost would be $35
Hi
Here is you answer mate
But don’t forget to mark me the brainliest
Plug the applicable numbers into the compound interest formula and see which is more.
A = p(1+r/n)nt
A = future amount
p =principal investment
r = interest rate as a decimal
n = number times compounded per year
t = time in years
A = 5000(1+.0743/365)365(10)
= 5000(1.000203562)3650 = $10,510.38
A = 5000(1+.075/4)4(10)
= 5000(1.01875)40 = $10,511.75
As you can see these are practically equal, but the 7.5% quarterly is more.
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
It will travel 180 miles
4.5 • 40= 180