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Trava [24]
2 years ago
15

Need help with this problem

Mathematics
2 answers:
netineya [11]2 years ago
8 0

Answer:


Step-by-step explanation:


alexandr1967 [171]2 years ago
4 0

Answer:

x = 10

Step-by-step explanation:

Since the triangles are similar the ratios of corresponding sides are equal

\frac{20}{20+8} = \frac{3x}{4x+2}, that is

\frac{20}{28} = \frac{3x}{4x+2} ( cross- multiply )

20(4x + 2) = 84x

80x + 40 = 84x ( subtract 80x from both sides )

40 = 4x ( divide both sides by 4 )

x = 10



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Step-by-step explanation:

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2 years ago
Find FG<br> FGH= 49<br> GH=31
Luba_88 [7]

Answer:

<h3>18</h3>

Step-by-step explanation:

From the given diagram;

FGH = 49

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Using the expression to get FG;

FG + GH = FGH

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4 0
2 years ago
Write a linear function from the table.
3241004551 [841]

Answer:

It would be A.

Step-by-step explanation:

Because, 5 x 0 is 0, 5 x 1 is 5, 5 x 2 is 10, and so on. Its really just counting in 10s.

3 0
2 years ago
businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message
KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let <em>X</em> = number of text messages receive or send in an hour.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: \lambda=\frac{62.7}{24}= 2.6125.

The probability of a random variable can be computed using the formula:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

             =1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

6 0
2 years ago
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