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2 years ago

The expression 8^x is equivalent to 32^y, where x and y are postive. what is the value of y/x

Mathematics

1 answer:

Daniel [21]2 years ago

6 0

Step-by-step explanation:

8^x = 2^3x

32^y = 2^5y

=> 3x = 5y

y/x = 3/5 (B)

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Find all critical numbers for the following function. Then use the second-derivative test on each critical number to determine

Gelneren [198K]

Critical values are values where f'(x)=0 and the bounds of a function. Thus, let's solve for f'(x)!

f(x)=2x^3-3x^2+3x+8
f'(x)=6x^2-6x+3

Now let's set f'(x)=0

0=6x^2-6x+3
0=2x^2-2x+1

As it turns out, 2x^2-2x+1 isn't factorable!
This saves me some time because this means there are no critical numbers!

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3 years ago

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BartSMP [9]

Answer:

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Step-by-step explanation:

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2 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

3 years ago

Solve for x. 2 ( x + 1 ) = 12

Evgen [1.6K]

2 ( x + 1 ) = 12

Use distributive property:

2x + 2 = 12

Subtract 2 from both sides:

2x = 10

Divide both sides by 2:

X = 5

5 0

3 years ago

Read 2 more answers

Answer plzz tell the truth

Tanzania [10]

Answer:

Is there any choices? I'd either say 3/5 or 60/100.

Step-by-step explanation:

The reason being, both would equal around 60% I believe. 60% is 3/5th's of a 100, and 60/100 is simply 60 out of 100. Unless your going farther than I'm not entirely sure. I'm sorry.

7 0

3 years ago

Read 2 more answers