1 - million
4 - hundred thousands
5 - ten thousands
3 - thousands
8 - hundreds
9 - tens
7 - ones
Threfore, the value of 4 is hundred thousands.
Answer:
f(x) = 3 if x ≤ -2
= 1 if x > -2 ⇒ attached figure
Step-by-step explanation:
* Lets explain how to answer the question
- For the part of the graph on the left side (2nd quadrant)
- There is a horizontal line start from x = -∞ and stop at x = -2
- The end of the line is black dot means x = -2 belongs to the function
- The horizontal line drawn at y = 3
∴ The equation of the horizontal line is y = 3
∴ The function represents this part of graph is y = 3 if x ≤ -2
- The other part of the graph is also horizontal line start from
x = -2 to x = ∞
- The end of the line is white dot means x = -2 does not belong
to the function
- The horizontal line drawn at y = 1
∴ The equation of the horizontal line is y = 1
∴ The function represents this part of graph is y = 1 if x > -2
* f(x) = 3 if x ≤ -2
= 1 if x > -2
- The answer is attached
Find the Z score of both:
Z =(X-μ)/σ
Z₁ = (60-80)/10 ==> -2
& Z₂ =(100-80)/10==>+2
& you will find on the Z score tables that the probability to be between
60 & 100 is 0.95 or 95%
Answer:
(a)![N(t)=Noe^{kt}](https://tex.z-dn.net/?f=N%28t%29%3DNoe%5E%7Bkt%7D)
(b)5,832 Mosquitoes
(c)5 days
Step-by-step explanation:
(a)Given an original amount
at t=0. The population of the colony with a growth rate
, where k is a constant is given as:
![N(t)=Noe^{kt}](https://tex.z-dn.net/?f=N%28t%29%3DNoe%5E%7Bkt%7D)
(b)If
and the population after 1 day, N(1)=1800
Then, from our model:
N(1)=1800
![1800=1000e^{k}\\$Divide both sides by 1000\\e^{k}=1.8\\$Take the natural logarithm of both sides\\k=ln(1.8)](https://tex.z-dn.net/?f=1800%3D1000e%5E%7Bk%7D%5C%5C%24Divide%20both%20sides%20by%201000%5C%5Ce%5E%7Bk%7D%3D1.8%5C%5C%24Take%20the%20natural%20logarithm%20of%20both%20sides%5C%5Ck%3Dln%281.8%29)
Therefore, our model is:
![N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t](https://tex.z-dn.net/?f=N%28t%29%3D1000e%5E%7Bt%2Aln%281.8%29%7D%5C%5CN%28t%29%3D1000%5Ccdot1.8%5Et)
In 3 days time
![N(3)=1000\cdot1.8^3=5832](https://tex.z-dn.net/?f=N%283%29%3D1000%5Ccdot1.8%5E3%3D5832)
The population of mosquitoes in 3 days time will be approximately 5832.
(c)If the population N(t)=20,000,we want to determine how many days it takes to attain that value.
From our model
![N(t)=1000\cdot1.8^t\\20000=1000\cdot1.8^t\\$Divide both sides by 1000\\20=1.8^t\\$Convert to logarithm form\\Log_{1.8}20=t\\\frac{Log 20}{Log 1.8}=t\\ t=5.097\approx 5\; days](https://tex.z-dn.net/?f=N%28t%29%3D1000%5Ccdot1.8%5Et%5C%5C20000%3D1000%5Ccdot1.8%5Et%5C%5C%24Divide%20both%20sides%20by%201000%5C%5C20%3D1.8%5Et%5C%5C%24Convert%20to%20logarithm%20form%5C%5CLog_%7B1.8%7D20%3Dt%5C%5C%5Cfrac%7BLog%2020%7D%7BLog%201.8%7D%3Dt%5C%5C%20t%3D5.097%5Capprox%205%5C%3B%20days)
In approximately 5 days, the population of mosquitoes will be 20,000.
Surfae area
we have 5 faces for the top shape
6 faces for the bottom
find areas of both
top=3*3+3*3+3*3+3*3=45
bottom one=3*10+3*3+3*3+3*10+3*10+3*(10-3)=129
add
45+129=174 mm^2
answer is B
volue
top shape is 3*3*3=27 mm^3
bottom is 10*3*3=90
add
27+90=117 mm^3
answer is C
6. 174 m^2 B
7. 117 mm^2 C