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2 years ago

Whats the Answer to this?

Mathematics

1 answer:

sineoko [7]2 years ago

7 0

Answer:

85%?

Step-by-step explanation:

going between so im guessing.

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Which shows only a vertical translation from the shaded figure onto the unshaded figure?

Gekata [30.6K]

Answer:

bruh u didnt even show the equation

Step-by-step explanation:

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2 years ago

Pens cost $0.20 each. Pencils cost $0.15 each. Write an algebraic expression that represents the total cost of buying p pens and

Harlamova29_29 [7]

Answer:

y=0.20p + 0.15n

Step-by-step explanation:

I believe this is correct I am 90% positive that it is right.

8 0

3 years ago

Jamal and Brianna each bought 3 oranges. They sliced each orange into 6 pieces. How many oranges slices did Jamal Brianna have a

SOVA2 [1]

Answer:The Answer is 18

Step-by-step explanation:if you have 3 oranges and each orange is sliced into 6 pieces each you multiply 3 by 6 and you get 18 and how you get 18 is you can add 6 three different times. :)

5 0

2 years ago

Identify the real and imaginary parts of the following complex number: 39i

Nataliya [291]

Answer:

An i is an imaginary value. So 7i is the imaginary part and 9 is the real part.

Hope this helps!!

3 0

2 years ago

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