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3 years ago

Question 5 (2) x cube - 3x square 9x - 5

1 answer:

3 0

Please, post the instructions along with your question, and if possible share the question in symbolic form, not in words.

Do you mean Question #5? By (2) x cube, do you mean 2x^3?

I strongly suggest that you use lots of parentheses ( ) to show how your numbers are grouped, and not to use " x " to denote multiplication (use " * " for that, please.

If only you'll clear this up, I'd be happy to help.

I will assume that your post is 2x^3 - 3(9x-5)^2.

Then 2x^3 - 3(81x - 90x + 25). Does this have any resemblance to what you wanted me to see in your post?

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There are 1,000 fish in a lake. Each year the population declines by 15%, but afterward, the lake is restocked with 500 aditiona

KatRina [158]

We begin with 1,000 fish in a lake.
Each year the population declines to 15%
Then they put back 500 fish in the lake.

The equation: (1000 * .85) +500

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2 years ago

What is. m + 0.6 = 2.6

professor190 [17]

Answer:

m=2

Step-by-step explanation:

STEP 1 : m+0.6=2.6

move constant to the right-hand side and change it's sign m=2.6-0.6

STEP 1: m=2.6-0.6

subtract the numbers m=2

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2 years ago

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2 years ago

Read 2 more answers

What is the slope of the line with equation 4x + 2y = 8?

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3 years ago

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A screening test for a certain disease is used in a large population of people of whom 1 in 1000 actually have the disease. Supp

sertanlavr [38]

Answer:

$P(D/T)=5.05*10^{-6}$

Step-by-step explanation:

Let's call D the event that a person has the disease, D' the event that a person doesn't have the disease and T the event that the person tests negative for the disease.

So, the probability P(D/T) that a randomly chosen person who tests negative for the disease actually has the disease is calculated as:

P(D/T) = P(D∩T)/P(T)

Where P(T) = P(D∩T) + P(D'∩T)

So, the probability P(D∩T) that a person has the disease and the person tests negative for the disease is equal to:

P(D∩T) = (1/1000)*(0.005) = 0.000005

Because 1/1000 is the probability that the person has the disease and 0.005 is the probability that the person tests negative given that the person has the disease.

At the same way, the probability P(D'∩T) that a person doesn't have the disease and the person tests negative for the disease is equal to:

P(D'∩T) = (999/1000)*(0.99) = 0.98901

Finally, P(T) and P(D/T) are equal to:

P(T) = 0.000005 + 0.98901 = 0.989015

$P(D/T) = 0.000005/0.989015 = 5.05*10^{-6}$

8 0

2 years ago