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netineya [11]
3 years ago
11

the ebola virus has an infection rate of 11% per day as compared to the SARS virus, which has a rate of 4% per day. If there wer

e one case of Ebola and 30 cases of SARS initially reported to the authorities and cases are reported each day, which statement is true?​

Mathematics
1 answer:
fomenos3 years ago
5 0

Answer:

3) At day 10 there are more SARS cases, but at day 53 there are more Ebola cases.

Step-by-step explanation:

First let's get our variables.

Ebola Rate = 11% or 0.11

SARS Rate = 4% or 0.04

x = number of days

Let Ebola = e(x)

Let SARS = s(x)

So the total number of cases  are:

e(x) = 1(1.11)^{x}

s(x) = 30(1.04)^{x}

At day 10:

e(10) = 1(1.11)^{10}

e(10) = 1(2.84)

e(10) = 2.84

s(10) = 30(1.04)^{10}

s(10) = 30(1.48)

s(10) = 44.41

So at day 10 the number of cases of Ebola will be 2 and SARS at 44.

At day 53:

e(53) = 1(1.11)^{53}

e(53) = 1(252.42)

e(53) = 252.42

s(53) = 30(1.04)^{53}

s(53) = 30(7.99)

s(53) = 239.82

By day 53, Ebola will overtake SARS at 252 cases to 239 cases.

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In a dog show, how many ways can 3 beagles, 2 poodles, and 5 grey hounds line up if the dogs of the same breed are considered id
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Complete the coordinate proof of the theorem.
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the length of AC can be calculated with the theorem of Pythagoras:

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seeing as the length of AC is the longest, it can be calculated by the following formula:

It is called "Pythagoras' Theorem" and can be written in one short equation:

a^2 + b^2 = c^2 (^ means to the power of by the way)

in this case, A and B are lengths AB and BC, so lenght AC can be calculated as the following:

a^2 + b^2 = (length AC)^2
length AC = √(a^2 + b^2)

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3 0
3 years ago
Read 2 more answers
Please Help. I really don’t get this concept, if you could explain it in detail it would be very appreciated
Dafna11 [192]
<h3>Answer: 24/25</h3>

=================================================

Explanation:

Sine is given to be negative, and so is tangent. This only happens in quadrant Q4

Recall that y = sin(theta), so if sin(theta) < 0, then we're below the x axis.

If tan(theta) < 0, then this means cos(theta) > 0

So we have y < 0 and x > 0 which places the angle somewhere in Q4.

--------------------------

Draw a right triangle as shown below in the attached image. We have AC = 25 and BC = 7. Use the pythagorean theorem to find that AB = 24

So this is what your steps may look like

a^2+b^2 = c^2

7^2+b^2 = 25^2

b^2+49 = 625

b^2 = 625-49

b^2 = 576

b = sqrt(576)

b = 24

So because AB = 24, we know that the cosine of the angle is adjacent/hypotenuse = 24/25

---------------------------

As an alternative, you could use the trig identity

sin^2(x) + cos^2(x) = 1

and plug in the given value of sine to solve for cosine. The cosine value result will be positive since we're in Q4.

So,

sin^2(x) + cos^2(x) = 1

(-7/25)^2 + cos^2(x) = 1

(49/625) + cos^2(x) = 1

cos^2(x) = 1 - (49/625)

cos^2(x) = (625/625) - (49/625)

cos^2(x) = (625-49)/625

cos^2(x) = 576/625

cos(x) = sqrt(576/625)

cos(x) = sqrt(576)/sqrt(625)

cos(x) = 24/25

This is effectively a rephrasing of the previous section since the pythagorean trig identity is more or less the pythagorean theorem (just in a trig form)

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Answer:

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