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Alenkasestr [34]

3 years ago

13

What is the prime fractorization of 84

2 answers:

Anvisha [2.4K]3 years ago

8 0

84/2 = 42
42/2 = 21
21/3 = 7
7/7 = 1

84 = 2^2 * 3 * 7

alexira [117]3 years ago

6 0

84 = 2 x 2 x 3 x 7 which can also be written 2^2x3x7.
1x84, 2x42, 3x28, 4x21, 6x14, or 7x12.

Hope this helped :)
Have a great day

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Which of the following expressions is correct?

RUDIKE [14]

<em>l-37l=37 is the correct answer.</em>

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3 years ago

You flip a coin 10 times and find the experimental probability of flipping tails to be 0.7. Does this seem reasonable? Explain.

Vedmedyk [2.9K]

No. I would say 50/50 chance. This answer probably isnt helpful :/

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3 years ago

Please help. It’s a picture

Korolek [52]

$260

I don’t know what model/formula you are supposed to be using.

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6 0

3 years ago

Solve the percent equation. 646 is what percent of 340?

ella [17]

646 is 190% of 340

Simply do the following,

1) divide,

$646 \div 340 = 1.9$

2) multiply the above answer by 100

$1.9 \times 100 = 190$

3) dock on that percent sign,

$190\%$

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I hope that helps you out!!

Any more questions, please just ask me!!

~ Zoey

4 0

3 years ago

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ. $e^{-\lambda.x}$

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - $e^{-\lambda.x}$

(a) <u>Probability</u> of distance at most <u>100m</u>, with λ = 0.0143:

F(100) = 1 - $e^{-0.0143.100}$

F(100) = 0.76

<u>Probability</u> of distance at most <u>200</u>:

F(200) = 1 - $e^{-0.0143.200}$

F(200) = 0.94

<u>Probability</u> of distance between <u>100 and 200</u>:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = $\frac{1}{\lambda}$

E(X) = $\frac{1}{0.0143}$

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = $\sqrt{\frac{1}{\lambda^{2}} }$

σ = $\sqrt{\frac{1}{0.0143^{2}} }$

σ = 69.93

<u>Distance exceeds the mean distance by more than 2σ</u>:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - $e^{-0.0143*209.79}$ )

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

$\int\limits^m_0 f({x}) \, dx$ = 0.5

$\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx$ = 0.5

$\lambda.\frac{e^{-\lambda.x}}{-\lambda}$ = $-e^{-\lambda.x} + e^{0}$

$1 - e^{-\lambda.m}$ = 0.5

$-e^{-\lambda.m}$ = - 0.5

ln( $e^{-0.0143.m}$ ) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0

4 years ago