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3 years ago

Y = -5x y = x-6 What is the answer

Mathematics

1 answer:

deff fn [24]3 years ago

8 0

Answer:

(1,-5)

Step-by-step explanation:

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Identify what Number property is shown for each statement

Alecsey [184]

i think answer are big enough

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3 years ago

Tax and tip word problems You might need: Calculator Problem Neal buys a board game. He pays for the board game and pays \$1.54$

Masja [62]

Answer:

$28

Step-by-step explanation:

SO,

5.5% OF SOME NUMBER is 1.54 dollars. Right?? Yes!

SOME NUMBER is what we are looking for, that is the ORIGINAL PRICE OF THE BOARD GAME.

We can let original price [some number] be "x" and write the word equation as algebraic equation shown below:

$0.055*x=1.54$

Note: 5.5% = 5.5/100 = 0.055

Now, we just simpyl solve for x:

$0.055*x=1.54\\x=\frac{1.54}{0.055}\\x=28$

$28 is the original price

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3 years ago

Please help it’s geometry will mark brainliest!

Dima020 [189]

Answer:

its 216

Step-by-step explanation:

when it's one the edge the arc is double the angle

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3 years ago

1. Solve for Y 1/2y+4x=z

Softa [21]

I hope this helps you

3 0

3 years ago

The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.4096, 0.4096, 0.1536, 0.0

Zinaida [17]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And if we replace we got:

$E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8$

So we expect about 0.8 defective computes in a batch of 4 selected.

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

Solution to the problem

For this case we have the following distribution given:

X 0 1 2 3 4

P(X) 0.4096 0.4096 0.1536 0.0256 0.0016

And we satisfy that $P(X_i) \geq 0$ and $\sum P(X_i) =1$ so we have a probability distribution. And we can find the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And if we replace we got:

$E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8$

So we expect about 0.8 defective computes in a batch of 4 selected.

5 0

4 years ago