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olya-2409 [2.1K]
3 years ago
14

Due to smoking a terminally ill patient loses approximately 2% of his lung capacity every 1 month at this rate begining at 100%

capacity what will the patients lung capacity be in one year (12 months) remeber the formula for exponential decay is y=a(1-r)^x
a. 69.4%
b. 76.0%
c. 88.6%
d. 78.5%
Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
3 0
I believe the answer is B.

Because the patient loses 2% every month. There’s 12 months in a year. 2 x 12 is 24. In a year he loses 24 percent, 100%- 24% = 76%. Which the answer the rest to be 76%
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The question is in the picture. PLEASE HELP!!!! Thank you!!!;);)
fomenos

Answer:

19/9

Step-by-step explanation:

16/27^2/3=19/9

hope this is helpful! brainly deleted my last response for some reason

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3 years ago
What is the correct operation to solve for x
FinnZ [79.3K]
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3 years ago
An online health and medicine company claims that about 50% of Internet users would go to an online sight first for
Serggg [28]

Answer:

The claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.

Step-by-step explanation:

Since an online health and medicine company claims that about 50% of Internet users would go to an online sight first for information regarding health and medicine, and a random sample of 1318 Internet users was asked where they will go for information the next time they need information about health or medicine; 606 said that they would use the Internet, to test the claim that less than 50% of Internet users get their health questions answered online, the following calculation must be performed:

1318 = 100

606 = X

606 x 100/1318 = X

60600/1318 = X

45.978 = X

Therefore, the claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.

6 0
3 years ago
Zaheer, a boy of height 1.5m was watching the entire programme. Initially he observed the top of the flag pole at an angle of el
7nadin3 [17]

Answer:

15.2 m

Step-by-step explanation:

You need to draw a figure. Start by drawing a horizontal segment approximately 10 cm long; that is the ground. Label the left end point A and the right endpoint B. On the right endpoint, B, go up a short 1 cm vertically. That is 1.5 m, the height of Zaheer. Label that point C. Now from that point draw a horizontal line that ends up above point A. Label that point D. Now go back to point C. Draw a segment up to the left at a 30 deg angle with CD. End the segment vertically above point D. Label that point E. That is the top of the flagpole. Draw a vertical segment down from point E through point D ending at point A. Segment AE is the flagpole. Go back to point C. Move 3 cm to the left on segment CD, and draw a point there and label it F. That is where Zaheer moved to. Now connect point F to point E. That is a 45-deg elevation to point E, the top of the flagpole.

m<EFD = 45 deg

m<EFC = 135 deg

m<FEC = 15 deg

m<ECD = 30 deg

We now use the law of sines to find EC

(sin 15)/10 = (sin 135)/EC

EC = 27.32

Because of the 30-60-90 triangle, ED = EC/2

ED = 13.66

Now we add the height of Zaheer to find AE.

13.66 + 1.5 = 15.16

Answer: 15.2 m

7 0
3 years ago
Find e^cos(2+3i) as a complex number expressed in Cartesian form.
ozzi

Answer:

The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2).

Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

Then,

\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

Now, recall the definition of the complex exponential:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

So,

e^{2i-3} = e^{-3}(\cos 2+i\sin 2)

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2.

Now, substituting this equality in (I) we get

\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2.

Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

5 0
3 years ago
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