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bearhunter [10]
3 years ago
9

A gas pump measures volume of gas to the nearest 0.01 gallon. Which measurement shows an appropriate level of precision for the

pump?
Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
5 0

Answer:

12.3 Gallons

Step-by-step explanation:

Options of the question:-

Option A. 12.33 gallons

Option B. 10 gallons

Option C. 12 gallons

Option D. 12.3 gallons

Option D is the answer :-

As the least count or zero error is 0.1 so it can measure only upto one decimal place so answer will be 12.3

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Luke plotted the number of hits he got in his baseball games this season versus the number of games played on a scatter plot. He
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Step-by-step explanation:

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3 years ago
Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.
Andre45 [30]

Answer:

Option d)  5 to the power of negative 5 over 6 is correct.

\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}

Above equation can be written as 5 to the power of negative 5 over 6.

ie, 5^\frac{\bf -5}{\bf 6}

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}

Above equation can be written as 5 to the power of negative 5 over 6.

7 0
3 years ago
Simplify each expression. 6mn3 -mn2 + 3mn3 +15mn2??
bekas [8.4K]

Answer:

9m(n)^3 +14m(n)^2

Step-by-step explanation:

6m(n)^3 - m(n)^2 + 3m(n)^3 + 15m(n)^2

=> 6m(n)^3 + 3m(n)^3 + 15m(n)^2 - m(n)^2

=> 9m(n)^3 +14m(n)^2

4 0
3 years ago
If N is the population of the colony and t is the time in​ days, express N as a function of t. Consider Upper N 0 is the origina
lesya [120]

Answer:

(a)N(t)=Noe^{kt}

(b)5,832 Mosquitoes

(c)5 days

Step-by-step explanation:

(a)Given an original amount N_o at t=0. The population of the colony with a growth rate k \neq 0, where k is a constant is given as:

N(t)=Noe^{kt}

(b)If N_o=1000 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)

Therefore, our model is:

N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t

In 3 days time

N(3)=1000\cdot1.8^3=5832

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

In approximately 5 days, the population of mosquitoes will be 20,000.

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