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Salsk061 [2.6K]

3 years ago

7

What is the value of x in the solution to the system of linear equations?

1 answer:

Romashka [77]3 years ago

6 0

Answer:

i think it is C

Step-by-step explanation

are the only equivalent answer to y and v

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Lines a and b are parallel<br> What is the value of x<br> 5<br> 10<br> 35<br> 55

arsen [322]

Answer:

5

Step-by-step explanation:

3 0

2 years ago

The length of 13 pink earasers lined up end to end is 29 1/4 inches. What would be the length of two earaser?

stepladder [879]

First you would want to find the length of one eraser and then multiply that by 2. To find the length of one you take the total length or all of them and divide it my the total number of them. 29 1/4 is 117/4.

(117/4) / 13 Dividing by something is the same as multiplying by the opposite.

117/4 × 1/13 = 117/52

Now you multiply by 2.

117/52 * 2/1 Cross simplify

117/26 * 1/1

117/26 inches or 4 1/2 inches.

7 0

3 years ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

a) $P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

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 successes

e = 2.71828 is the Euler number

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

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

So

$P(X \geq 2) = 0.93803$

5 0

2 years ago

If an x-coordinate repeats, can a relation still be a function?<br><br>I need help right now

Gre4nikov [31]

Answer:

No

Step-by-step explanation:

if you see any duplicates or repetitions in the x-values, the relation is not a function.

8 0

2 years ago

Which answer choice below represents the greatest percent change? F. To earn a position on the school’s wrestling team, Colin in

julsineya [31]

I believe it to be F but i’m not sure

6 0

2 years ago

Read 2 more answers