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vazorg [7]
2 years ago
8

On a 1:5 scale drawing of a car one part is 8 inches how long will the actual car part be

Mathematics
2 answers:
Lemur [1.5K]2 years ago
6 0
40


I hope this helps and have a wonderful day filled with joy!!
lina2011 [118]2 years ago
5 0
40, because if the ratio is 1:5 then we can times 8 by 5 to get 40 (I think :P)
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2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA
professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

A=45e^{-0.0045(t)}

a) To find the initial amount of this substance

At t=0, we get

A=45e^{-0.0045(0)}A=45e^0

We know that e^0=1 ( anything to the power zero is 1)

we get,

A=45

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

\frac{45}{2}=45e^{-0.0045(t)}

\frac{1}{2}=e^{-0.0045(t)}

Taking natural logarithm on both sides we get,

\ln (\frac{1}{2})=-0.0045(t)^{}(-1)\ln (\frac{1}{2})=0.0045(t)\ln (\frac{1}{2})^{-1}=0.0045(t)\ln (2)=0.0045(t)0.6931=0.0045(t)t=\frac{0.6931}{0.0045}t=154.02

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

A=45e^{-0.0045(2500)}A=45e^{-11.25}A=45\times0.000013=0.000585A=0.000585

The amount of substance will be present around in 2500 years is 0.000585 grams

4 0
8 months ago
NEED HELP FAST WILL GIVE BRAILYIST What is the solution to Negative 1 minus 7?<br><br> –8 –6 6 8
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Answer:

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The experimental probability of seeing a hawk at the Avian Viewing Center on any given day is 20%. If Allison visits the center
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The experimental probability is, as the name suggests, a probability based on observation. If we say that the experimental probability of seeing a hawk at the Avian Viewing Center on any given day is 20%, it means that someone has visited the center for many days, and at the end of this experiment he has met hawks on 20% of the days, i.e. one out of five days.

If we assume that these measurement are trustworthy, we can assume that Allison will also see a hawk on one fifth of the days.

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\cfrac{240}{5} = 48 days.

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