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3 years ago

How does the product 110×any number compare to the product 11× any number

Mathematics

1 answer:

Effectus [21]3 years ago

6 0

The product of 100 x a number would be 10 times larger than the product of 11 x a number

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First piece is 5.2 cm the second piece must be 1.5 times at long as the first

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What do u need to do? add, subtract, multiply, or divide?

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riadik2000 [5.3K]

Answer:

$x=20; y=50$ or $(20, 50)$

Step-by-step explanation:

Substitution means plugging in one variable's value that consists of the opposite variable. Because $y=3x-10$ is already specified, you can plug it into the second equation's $y$ value. After doing that, it looks like this:

$-4x+2y=20→-4x+2(3x-10)=20$

Then you would distribute the $3$ across the parentheses next to it, like this:

$-4x+2(3x-10)=20→-4x+6x-20=20$

Then, add like terms, like this:

$-4x+6x-20=20→2x=40$

Then divide both side by $2$ to isolate $x$ .

$x=20$

Now, you can plug $20$ ( $x$ ) into either equation, but the first one seems simpler so you would pick that. It would look like this:

$y=3x-10→y=3(20)-10$

Solving would look like this:

$y=3(20)-10→y=60-10→y=50$

So the answer is $x=20; y=50$ or $(20, 50)$ .

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3 years ago

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What is four points fewer than the bulls scored write in a algebraic expression

Elena-2011 [213]

B - 4 where B is the variable representing the Bulls score

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3 years ago

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What is the equation of the line that passes through the point (−3,−11) and is parallel to the line with the equation y=2x−4

tester [92]

parallel means same slope. m = 2

y - y₁ = m(x - x₁)

y - (-11)= 2(x - (-3))

y + 11 = 2(x + 3)

y + 11 = 2x + 6

y = 2x - 5

Answer: A

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3 years ago

Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo

natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

P(B) is the probability of the person having cancer, so $P(B) = 0.009$

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So $P(A/B) = 0.91$

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

$P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765$

What is the probability that the person actually does have cancer?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213$

There is a 12.13% probability that the person actually does have cancer.

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3 years ago