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3 years ago

7

What is the constant of proportionality in the equation StartFraction x over y EndFraction = StartFraction 2 over 9 EndFraction?

HELP PLZ I WILL NAME BRAINLEST

1 answer:

ratelena [41]3 years ago

3 0

Answer:

ITS 5/4

Step-by-step explanation:

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{1, 2, 6, 24, 120, ...}

enyata [817]

This is a factorial sequence that can be modeled by An = n!. As you may see, the increasing numbers are factorials of 1,2,3,4, and 5. Factorial means multiplying backwards and is represented by !. For example, 1! is 1*1 =1. 2! is 2*1, 3! is 3*2*1, 4! is 4*3*2*1 etc.

8 0

3 years ago

Jim spent $46.16 on some mulch and one shovel. Each bag of mulch costs $3.49 and the shovel cost $14.75. How many bags of mulch

mezya [45]

To find the amount he spent on bags of mulch you must take the cost of the shovel away from the total, giving the equation:
46.16-14.75=3.49x
To covert this to whole numbers, multiply by 100
4616-1475=349x
3141=349x
x=9

4 0

4 years ago

Which expression is equivalent to (x Superscript one-half Baseline y Superscript negative one-fourth Baseline z) Superscript neg

Darina [25.2K]

By using exponent properties, we will get the simplified expression:

$x^{-1}*y^{1/2}*z^2$

<h3>How to simplify the given expression?</h3>

Here we have the expression:

$(x^{1/2}*y^{-1/4}*z)^{-2}$

Remember the exponent properties:

$(a^n)^m = a^{n*m}$

And:

$(a*b)^n = (a^n)*(b^n)$

So using these two properties, we can rewrite:

$(x^{1/2})^{-2}*(y^{-1/4})^{-2}*(z)^{-2}\\\\(x^{-2*1/2})*(y^{-2*-1/4})*(z^{-2}})\\\\x^{-1}*y^{1/2}*z^2$

So we conclude that the completely simplified expression is:

$x^{-1}*y^{1/2}*z^2$

If you want to learn more about exponents:

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8 0

2 years ago

Select the equation written in point-slope form.

slavikrds [6]

Poin slope form is
y-y1=m(x-1)
the answer is A

4 0

3 years ago

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o

timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

$\lambda t=8\times 1=8$

The probability distribution of a Poisson distribution is:

$P(X=x)=\frac{e^{-8}(8)^{x}}{x!}$

Compute the value of P (X = 6) as follows:

$P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214$

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

$P(X\geq 6)=1-P(X$

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

$P(X\geq 10)=1-P(X$

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

$\lambda t=8\times 1.5=12$

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

$SD=\sqrt{\lambda t}=\sqrt{12}=3.464$

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

$\lambda t=8\times 2.5=20$

Compute the value of P (X ≥ 20) as follows:

$P(X\geq 20)=1-P(X$

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

$P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108$

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0

3 years ago