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Musya8 [376]
1 year ago
5

Of the 125 members of the marching band, 42 are in chorus, 78 play

Mathematics
1 answer:
Margarita [4]1 year ago
4 0

Using it's concept, it is found that there is a 0.2979 = 29.79% probability that they are in chorus, given that they do not play a sport.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem, 125 - 78 = 47 members of the band do not play a sport, of which 42 - 28 = 14 are in the chorus, hence the probability is given by:

p = 14/47 = 0.2979.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

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It is 5/1
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3 years ago
An initial of the $100 is now valuated at $150. The annual interest rate is 5%, compounded continuously. The equation 100e^0.05=
tangare [24]

Answer:

8 years approximately

The problem (I'm assuming is):

Solve 100e^{0.05t}=150.

I put a t in the problem where I suspected it in went.

Step-by-step explanation:

100e^{0.05t}=150

Divide both sides by 100:

e^{0.05t}=\frac{150}{100}

e^{0.05t}=1.5

Rewrite in logarithm form.

\ln(1.5)=0.05t

Divide both sides by 0.05:

\frac{\ln(1.5)}{0.05}=t

Put left hand side into calculator:

8.109 \approx t

t \approx 8.109

So about 8 years if I wrote down the equation correctly.

4 0
3 years ago
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oksian1 [2.3K]
Subtract 15 from both sides

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3 years ago
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(xy^-6)^2<br><br>A. xy^-36<br>B. xy^36<br>C. x^2y^-12​
Makovka662 [10]

Answer:

Step-by-step explanation:       (9=4-38-838,256=2-60^7

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Use​ Newton's method to find an approximate solution of ln(x)=10-x. Start with x_0 =9 and find x_2 .
vova2212 [387]

Answer:

x₂ = 7.9156

Step-by-step explanation:

Given the function  ln(x)=10-x with initial value x₀ = 9, we are to find the second approximation value x₂ using the Newton's method. According to Newtons method xₙ₊₁ = xₙ -  f(xₙ)/f'(xₙ)

If f(x) = ln(x)+x-10

f'(x) = 1/x + 1

f(9) = ln9+9-10

f(9) = ln9- 1

f(9) = 2.1972 - 1

f(9) = 1.1972

f'(9) = 1/9 + 1

f'(9) = 10/9

f'(9) = 1.1111

x₁ = x₀ -  f(x₀)/f'(x₀)

x₁ = 9 -  1.1972/1.1111

x₁  = 9 - 1.0775

x₁  = 7.9225

x₂ = x₁ -  f(x₁)/f'(x₁)

x₂ = 7.9225 -  f(7.9225)/f'(7.9225)

f(7.9225) = ln7.9225 + 7.9225 -10

f(7.9225) = 2.0697 + 7.9225 -10

f(7.9225) = 0.0078

f'(7.9225) = 1/7.9225 + 1

f'(7.9225) = 0.1262+1

f'(7.9225) = 1.1262

x₂ = 7.9225 - 0.0078/1.1262

x₂ = 7.9225 - 0.006926

x₂ = 7.9156

<em>Hence the approximate value of x₂ is 7.9156</em>

7 0
3 years ago
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