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3 years ago

What would the coordinate be 2x+8y=20 y=2

Mathematics

2 answers:

evablogger [386]3 years ago

7 0

Since y=2,

We have 2x+8(2)=20

Therefore, 2x=4

X=2

So the coordinate would be ( 2, 2 )

svetlana [45]3 years ago

6 0

2x+8y=20;y=2
2x+8(2)=20
2x+16=20
-16 -16
-------------------
2x=4
x=2
The coordinate will be (2,2)! Hope that was helpful :)

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PLEASE HELP!!!! Geometry is so hard and this is due tonight :(

omeli [17]

Cos(68)=z/24
Z=24cos(68)
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X=cos^-1(16/34)
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7 0

3 years ago

2x +4=12 step by step

Alekssandra [29.7K]

x=4

Step-by-step explanation:

2x+4=12

-4 -4

---------------

2x = 8

---- ---

2 2

x=4

6 0

3 years ago

Let f(x) = -5x - 4 and g(x) = 6x - 7. Find f(x) + g(x)

Fynjy0 [20]

Simple...

if f(x)= -5x-4 and g(x)=6x-7

And you want f(x)+g(x)

then just add.....f(x)+g(x)

-5x-4+6x-7

1x-11

This is the answer to f(x)+g(x)=1x-11

Thus, your answer.

3 0

3 years ago

5.53 An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day. What

Licemer1 [7]

Answer:

a) 38.4% probability that on a given day this item is requested more than 5 times.

b) 0.67% probability that on a given day this item is not requested at all.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day.

This means that $\mu = 5$

What is the probability that on a given day this item is requested

(a) more than 5 times?

Either it is requested 5 times or less, or it is requested more than 5 times. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067$

$P(X = 1) = \frac{e^{-5}*(5)^{1}}{(1)!} = 0.0337$

$P(X = 2) = \frac{e^{-5}*(5)^{2}}{(2)!} = 0.0842$

$P(X = 3) = \frac{e^{-5}*(5)^{3}}{(3)!} = 0.1404$

$P(X = 4) = \frac{e^{-5}*(5)^{4}}{(4)!} = 0.1755$

$P(X = 5) = \frac{e^{-5}*(5)^{5}}{(5)!} = 0.1755$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 + 0.1755 = 0.616$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.616 = 0.384$

38.4% probability that on a given day this item is requested more than 5 times.

(b) not at all?

$P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067$

0.67% probability that on a given day this item is not requested at all.

3 0

3 years ago

Find the area of the triangle

Allushta [10]

Answer:

75.03

Step-by-step explanation:

14×11×sin(77)

= 75.02649499..

≈ 75.03 (rounded to nearest hundredth)

Answered by GAUTHMATH

7 0

3 years ago