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Elodia [21]
2 years ago
5

Please help meeeeee, timed.

Mathematics
1 answer:
zepelin [54]2 years ago
4 0
Answer C: no solution
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The yearbook club is handing out T-shirts to its members. There are 5 blue, 7 green, 9 red, and 4 yellow T-shirts in all. If Jac
damaskus [11]

Answer:

n(R)=9

n(Y)=4

n(B)=5

n(G)=7

n(s)25

probability of getting red =n(R)/n(S)=9/25

6 0
3 years ago
Find all solutions to the equation in the interval [0, 2pi). COS X = sin 2x​
Lera25 [3.4K]

Answer:

Step-by-step explanation:

cos x=sin 2x

sin 2x-cos x=0

2 sin x cos x-cos x=0

cos x(2 sin x-1)=0

either cos x=0

x=\frac{\pi }{2},\frac{3\pi }{2}

or

2 sin x-1=0

sin x=\frac{1}{2} =sin (\frac{\pi }{6}),sin(\pi -\frac{\pi }{6} )\\x=\frac{\pi }{6} ,\frac{5 \pi }{6}

6 0
3 years ago
What is the area of the composite figure?( 20 points)​
kicyunya [14]
The answer is B. 20.285
5 0
3 years ago
Algebra Question ( Matrices and Determinants ) 20 point
blagie [28]
The determinant of a 2 x 2 matrix can be calculated as:
Product of non-diagonal elements subtracted from product of diagonal elements.

The diagonal elements in given matrix are 12 and 2. The non-diagonal elements are -6 and 0.

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Determinant G = 12(2) - (-6)(0)

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6 0
3 years ago
Read 2 more answers
The probability that a rental car will be stolen is. 4. if 3500 cars are rented, what is the approximate poisson probability tha
kozerog [31]

Using the Poisson distribution, there is a 0.8335 = 83.35% probability that 2 or fewer will be stolen.

<h3>What is the Poisson distribution?</h3>

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

The parameters are:

  • x is the number of successes
  • e = 2.71828 is the Euler number
  • \mu is the mean in the given interval.

The probability that a rental car will be stolen is 0.0004, hence, for 3500 cars, the mean is:

\mu = 3500 \times 0.0004 = 1.4

The probability that 2 or fewer cars will be stolen is:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-1.4}1.4^{0}}{(0)!} = 0.2466

P(X = 1) = \frac{e^{-1.4}1.4^{1}}{(1)!} = 0.3452

P(X = 2) = \frac{e^{-1.4}1.4^{2}}{(2)!} = 0.2417

Then:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2466 + 0.3452 + 0.2417 = 0.8335

0.8335 = 83.35% probability that 2 or fewer will be stolen.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

4 0
1 year ago
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