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2 years ago

Due in 20 mins help pls

Mathematics

1 answer:

kodGreya [7K]2 years ago

6 0

Answer: ....

P/s:

2 × 3/12 = 3 × 2/12 = 6 × 1/12 (= 1/2)
4 × 3/12 = 1

ok done. Thank to me :>

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Is this a funcion? Explain why.<br><br> X: -2, 0, 1, -2<br> Y:4, 5, 4, 5

Alina [70]

Answer:

Step-by-step explanation:

it isn't a Function because the Domain (X coordinate) repeats a number.

Take this note:

Domain (x) repeats: Not a function

Domain(x) doesn't repeat: Function

Hope this helps ;)

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2 years ago

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Please help quickly thanks :)

marusya05 [52]

(0, 0) → t = 0, w = 0 → w = 0 · 0

(1, 4) → t = 1, w = 4 → w = 1 · 4

(2, 8) → t = 2, w = 8 → w = 2 · 4

(3, 12) → t = 3, w = 12 → w = 3 · 4

<h3>Answer: w = 4t</h3>

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3 years ago

Find the distance between the two points in simplest radical form for (-6,1) and (-8,-4)

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Answer:

Step-by-step explanation:

Δy =-4-1 = -5

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3 years ago

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Solve for x.

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X^3 = 216

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3 years ago

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It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha

stira [4]

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

0.54% of the calls receive a busy signal, hence p = 0.0054.

A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

$P(X \geq 5) = 1 - P(X < 5)$

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009$

$P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062$

$P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218$

$P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513$

$P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903$

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295$

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

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2 years ago