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NeTakaya
3 years ago
9

you send out 20,000 emails of 6% are opened of those 9% clicked on the link to register for something of those who clicked on th

e link 30% complete the registration how many people completed the registration​
Mathematics
2 answers:
nikitadnepr [17]3 years ago
6 0

Answer:

<u>Rounding to the next whole, 33 persons completed the registration</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of e-mails sent out = 20,000

Number of e-mails opened = 6%

Number of e-mails that clicked on the link to register = 9% of the 6%

Number of e-mails that completed the registration = 30% of 9% of 6%

2.  How many people completed the registration​?

Let's do all the calculations to answer the question, this way:

Number of e-mails opened = 6 % of 20,000

Number of e-mails opened =0.06 * 20,000 = 1,200

Number of e-mails that clicked on the link to register = 9% of the 6%

Number of e-mails that clicked on the link to register = 0.09 * 1,200 = 108

Number of e-mails that completed the registration = 30% of 9% of 6%

Number of e-mails that completed the registration = 0.3 * 108 = 32.4

<u>Rounding to the next whole, 33 persons completed the registration</u>

maxonik [38]3 years ago
4 0

Answer:

The answer is 32.

Step-by-step explanation

hope this helps :D

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3 years ago
Angle 7 and angle 8 are complementary. angle 5 equals angle 8 and angle 6=29. ​
olya-2409 [2.1K]

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3 years ago
What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)
Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\

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Again repeating the same procedure we get

=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\

Again repeating the same procedure we get

\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\

Now solving this integral we have

\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}

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L[7t^{3}]=\frac{7\times 3\times 2}{s^4}

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Step-by-step explanation:

hope it helps!

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