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3 years ago

Solve the equation for the given replacement set. State the solution set.

1 answer:

3 0

X=3 . ok well ur gonna do reverse mathematics so start by subtracting 2 from 17 (bc the oppistite of addition is subtraction) an then rewrite the equation. 5x=15 and then divide 15 by 5 and it equals 3 and then do it all over again meaning put the equation as it was but with the value. 5x3+2=17 and it all makes sense. im so sorry i answered so late i just got this question recommended to me.

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Use the Phythagorean theorem to find the length of the third side

Brums [2.3K]

Answer:

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Step-by-step explanation:

by using pythagorus theorem,

=》

${x}^{2} = {7}^{2} - {5}^{2}$

=》

${x}^{2} = 49 - 25$

=》

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3 years ago

How much water will it take to fill the aquarium?

Soloha48 [4]

Answer:

4096

Step-by-step explanation:

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4 years ago

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Simplify it using order of operations

natta225 [31]

Answer:

=106

Step-by-step explanation:

PMDAS

First thing you do is the parenthesis

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then the exponent

6+4(25)

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3 years ago

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To convert 58.58 to a fraction, we evaluate 58.58-0.58 What is 58.58-0.58 ?

lord [1]

Answer:

Step-by-step explanation:

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4 years ago

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Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).

$P(Y=0) + P(Y \geq 1) = 1 \implies P(Y \geq 1) = 1 -P(Y=0) = 1 - 0.0821 = 0.918$

3 0

3 years ago