<img src="https://tex.z-dn.net/?f=%5Cfrac%7B12%7D%7B20%7D" id="TexFormula1" title="\frac{12}{20}" alt="\frac{12}{20}" align="abs

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Nonamiya [84]

1 year ago

middle" class="latex-formula"> = ______% = ______ hundredths

Mathematics

1 answer:

Thepotemich [5.8K]1 year ago

8 0

The percentage form of given fraction is 60% and the hundredths form is 0.60

According to the statement

we have given that the a fraction and we have to find the percentage of that fraction and write in the form hundredths.

So, For this purpose,

The given fraction is 12/20.

Then the definition of the percentage is that

The Percentage, a relative value indicating hundredth parts of any quantity.

so, the percentage of given fraction is :

Percentage fraction = 12/20 * 100

After solving it, The percentage fraction will become:

Percentage fraction = 60%

and Now convert into the hundredths form then

In the hundredths form it will become

from 60% to 0.60.

So, The percentage form of given fraction is 60% and the hundredths form is 0.60

Learn more about Percentage here

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Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .

c is a time time between 2:00 and 2:20 at which the acceleration is $60 \:{\frac{mi}{h^2}}$ .

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