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3 years ago

A(-2, 1), B(-4,-1), and C(0, -1) and is

1 answer:

5 0

(4,-1)

If you graph 0,-1 as a coordinate then x=2 you can just count how many are before the line, then count that many after and that’s your coordinate

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Select all ratios equivalent to 12 :8 15:10 5:3 25:16

lora16 [44]

Answer:

15:10

Step-by-step explanation:

any ratio that has an equivalent of 2/3 is equivalent to the ratio 12:8

3 0

3 years ago

Factories A, B and C produce computers. Factory A produces 4 times as manycomputers as factory C, and factory B produces 7 times

Elina [12.6K]

Answer:

The probability is $P(A') = 0.485$

Step-by-step explanation:

Let assume that the number of computer produced by factory C is k = 1

So From the question we are told that

The number produced by factory A is 4k = 4

The number produced by factory B is 7k = 7

The probability of defective computers from A is $P(A) = 0.04$

The probability of defective computers from B is $P(B) = 0.02$

The probability of defective computers from C is $P(C) = 0.03$

Now the probability of factory A producing a defective computer out of the 4 computers produced is

$P(a) = 4 * P(A)$

substituting values

$P(a) = 4 * 0.04$

$P(a) = 0.16$

The probability of factory B producing a defective computer out of the 7 computers produced is

$P(b) = 7 * P(B)$

substituting values

$P(b) = 7 * 0.02$

$P(b) = 0.14$

The probability of factory C producing a defective computer out of the 1 computer produced is

$P(c) = 1 * P(C)$

substituting values

$P(c) = 1 * 0.03$

$P(b) = 0.03$

So the probability that the a computer produced from the three factory will be defective is

$P(t) = P(a) + P(b) + P(c)$

substituting values

$P(t) = 0.16 + 0.14 + 0.03$

$P(t) = 0.33$

Now the probability that the defective computer is produced from factory A is

$P(A') = \frac{P(a)}{P(t)}$

$P(A') = \frac{ 0.16}{0.33}$

$P(A') = 0.485$

3 0

4 years ago

Alexis has a pocket full of coins that consists of dimes and nickels. She has a total of 40 coins. The total value of her coins

MrMuchimi

Answer

3 quarters 7 dimes

Step-by-step explanation:

4 0

3 years ago

Given a function <img src="https://tex.z-dn.net/?f=f%28x%29%3D3x%5E4-5x%5E2%2B2x-3" id="TexFormula1" title="f(x)=3x^4-5x^2+2x-3"

Levart [38]

Answer:

$\huge\boxed{f(-1) = -7}$

Step-by-step explanation:

In order to solve for this function, we need to substitute in our value of x inside to find f(x). Since we are trying to evalue f(-1), we will substitute -1 in as x to our equation.

$f(-1) = 3(-1)^4 - 5(-1)^2 + 2(-1) - 3$

Now we can solve for the function by multiplying/subtracting/adding our known values.

Starting with the first term to the last term:

$3(-1)^4 = 3$

WAIT! How is this possible? $-1^4 = -1$ (according to my calculator), and $3 \cdot -1 = -3$ , not 3!

It's important to note that taking a power of a negative number and multiplying a negative number are two different things. Let's use $-2^2$ as an example.

What your calculator did was follow BEMDAS since it wasn't explicitly told not to.

BEMDAS:

- Brackets

- Exponents

- Multiplication/Division

- Addition/Subtraction

Examining the equation, your calculator used this rule properly. Note that exponents come over multiplication.

So rather than being "-2 squared" - it's "the negative of of 2 squared."

Tying this back into our problem, the squared method would only be true if it looks like $-1^4$ . However, since we're substituting in -1, it looks like $(-1)^4$ , so the expression reads out as "-1 to the fourth."