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3 years ago

What is 11kx+13kx=6 solve for the x

Mathematics

1 answer:

Art [367]3 years ago

3 0

11kx + 13kx = 6
24kx = 6
x = 6 / 24k = 1 / 4k Answer

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Help please True or false 15=11m. <br><br> if m=4?

DedPeter [7]

Answer:

False

Step-by-step explanation:

If F=4 that means 11 x 4 = 44 and 15 does not equal 44.

6 0

3 years ago

$22,000 divided by 18%

slavikrds [6]

18% = 18/100 = 0,18

22000/0,18 = 22000/(18/100) = (22000 . 100)/18 = 1100000/9 = 122222,2222...

So $22,000 divided by 18% is equal to $122.222,22

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3 years ago

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4 years ago

4. How many seconds will it take an echo to reach your ears if you yell toward a mountain

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3 years ago

3.- In a certain desert region the average number of persons who become seriously ill each year from eating a certain poisonous

Naily [24]

Answer:

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

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662$

$P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634$

$P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029$

$P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595$

$P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151$

And replacing we got:

$P(X \geq 5) =0.76493$

Step-by-step explanation:

Previous concepts

Let X the random variable that represent the number of people that will become sereiosly ill in two years. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this case the value for $\lambda$ would be:

$\lambda = 3.2 \frac{ills}{year} *2 years = 6.4$

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

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

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662$

$P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634$

$P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029$

$P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595$

$P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151$

And replacing we got:

$P(X \geq 5) =0.76493$

7 0

3 years ago

Read 2 more answers