The formula is y=0.5m+20
You start (0,20) then for each mile you increase by 0.5.
Or
Since the x axis is by two, then every two miles you increase by 1 mile. So (0,20); (2,21); (4,22); (6,23); (8,24).
Answer:
$3100 is invested at 9%
$4900 is invested at 11%
Step-by-step explanation:
Let's take "x" be the amount invested at 9%.
(x + 1800) is invested in another account at 11%.
The interest amount earned by the two accounts is $818.
Here we can use the simple interest formula and find the amount invested in each account.
Simple interest (I) = , where P- is the principal , N is the number of years and R is the interest rate.
Simple interest =
0.09x + 0.11(x+1800) = 818
Now we have to simplify and find the value of x .
Use the distributive property and simplify the second term.
0.09x + 0.11x + 198 = 818
0.2x + 198 = 818
0.2x =818 - 198
0.2x = 620
x = 620/0.2
x = 3100.
So $3100 is invested at 9%
x + 1800 = 3100 + 1800
= $4900
$4900 is invested at 11%
Hope this helped.
0.8=0.8/1=8/10=4/5
not one of them
but,
2/9=0.2222222222
4/9=0.444444444444
7/9=0.7777777777
8/9=0.88888888888
I think you meant 0.8 repeating so
answer is D
The height of an interior residential door is about 7 ft.
Doors on commercial buildings or upscale residences may be taller. The usual width is about 3 ft.
_____
My front door measures 6 2/3 ft high by about 3 ft wide. My bedroom doors are 2 1/2 ft wide. The minimum for ADA-compliant doors is 2 2/3 ft.
Answer:
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.
Step-by-step explanation:
We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.
The probability of k online retail orders that turn out to be fraudulent in the sample is:
We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:
![P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483](https://tex.z-dn.net/?f=P%28x%5Cgeq2%29%3D1-%5BP%28x%3D0%29%2BP%28x%3D1%29%5D%5C%5C%5C%5C%5C%5CP%28x%3D0%29%3D%5Cdbinom%7B20%7D%7B0%7D%5Ccdot0.08%5E%7B0%7D%5Ccdot0.92%5E%7B20%7D%3D1%5Ccdot1%5Ccdot0.189%3D0.189%5C%5C%5C%5C%5C%5CP%28x%3D1%29%3D%5Cdbinom%7B20%7D%7B1%7D%5Ccdot0.08%5E%7B1%7D%5Ccdot0.92%5E%7B19%7D%3D20%5Ccdot0.08%5Ccdot0.205%3D0.328%5C%5C%5C%5C%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-%5B0.189%2B0.328%5D%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-0.517%3D0.483)
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.
