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nadezda [96]
2 years ago
15

Kari is flying a kite.she realeses 50feet of string.what is the approximate difference in the height of the kite when the string

makes a 25 degree angle with the ground and when the string makes a 45 degree angle with the ground?round to the nearest tenth
Mathematics
1 answer:
Elza [17]2 years ago
4 0
Sinα=h/s

h=ssinα

h=50sinα

dh=h(45)-h(25)

dh=50(sin45-sin25) feet

dh≈14.2 ft (to nearest tenth of a foot)
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Even though the elimination method is easy I still have trouble with it
kicyunya [14]

Answer:

(1,2)

Step-by-step explanation:

x+4y = 9

2x -4y= -6

Add the equations together

x+4y = 9

2x -4y= -6

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3x +0y = 3

3x=3

Divide by 3

3x/3 = 3/3

x=1

Now find y

x+4y = 9

1 +4y =9

Subtract 1 from each side

4y = 8

Divide by 4

4y/4 = 8/4

y =2

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3 years ago
Express your final answer in simplest form 3/5 + -7/8
frez [133]

Answer:

- 0.5

Step-by-step explanation:

you first change the fractions numbers  in to decimal form

      so;   2/5 = 0.4

               -7/8 = - 0.875

hence;   0.4 + (-0.875)

             = - 0.475

in to its simplest form   =  - 0.5

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5 0
2 years ago
Solve x2 + 14x = −24 by completing the square.<br><br> What is the solution set of the equation?
yan [13]
Hope this helps with the question :)

3 0
3 years ago
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Step-by-step explanation:

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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
3 years ago
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