Answer:
400 miles per hour
Step-by-step explanation:
2.000 miles : 5 hours =400 miles per hour
have a nice day :)
Answer:
3.5 (thousand) boxes
Step-by-step explanation:
The line of best fit is the line that best fits the data / points. This line is already graphed for us, nice!
To find how much we should purchase if we only want to buy $5.05, we look to the y-axis (cost of each box of paper) and locate 5.05, in between the 5 and the 5.1. We look across to the x-axis (number of boxes) and see the answer is 3.5.
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
2(4x+5)>7x+20 perform indicated multiplication on left side
8x+10>7x+20 subtract 7x from both sides
x+10>20 subtract 10 from both sides
x>10
or in interval notation, x=(10, +oo)
Answer:
1 cm = 100 mi
28.9/1=28.9
28.9 x 100 = 2890 miles distance
2890 / 65 miles per hour = 44.4615 hours to drive
44.461/24 hours in a day = 1.85 rounded up is 2 days
About two days
If you have any questions you can ask
Step-by-step explanation:
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.
