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2 years ago

Please help with this?

Mathematics

1 answer:

Semmy [17]2 years ago

4 0

Answer: Subtract 3 from each side of the equation(A)

Step-by-step explanation:

3n2−15n=3

1:Subtract 3n2 from both sides.

-15n=3-3n2

2:Divide both sides by -15.

-15n/-15=3-3n2/-15

3:Dividing by −15 undoes the multiplication by −15.

n=3-3n2/-15

4:Divide 3−3n 2 by −15

n=n2-1/5

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hawnda incorrectly found the distance between -26 and 25 by following these steps: Step 1: |25 – (-26)| Step 2: |25 – 26| Step 3

tekilochka [14]

Answer:

Step 2

Step-by-step explanation:

Step 2 is not correct. It should be abs(25 - - 26) = abs(25 + 26) = 51

Given that Step 2 is accepted (as - 1) then three is OK and and so is 4. But technically they are incorrect as well. I think you are likely to say that step 2 is the problem.

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3 years ago

What 2 numbers whose sum is 105.7 the numbers are 19, 85.2, 533, 571, 88.2, 525, 20, 17.5, 400, 261, 20.5 ,125, 7, 23, 901, 30

givi [52]

To make it quickly we can consider only those numbers which will give us a decimal of xxx.7. There are only a few numbers like this: 85.2, 88.2, 20.5.

I tried adding 85.2 to 17.5 and it gave me 102.5 - it was close but not close enough :)

So I picked 85.2 + 20.5 = 105.7 and voila :)

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3 years ago

One-Step Equations:

erastova [34]

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3 years ago

Read 2 more answers

Verify that the following function is a probability mass function, and determine the requested probabilities. f left-parenthesis

Licemer1 [7]

Answer:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

Step-by-step explanation:

For this case we have the following density function:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

In order to satisfty that this function is a probability mass function we need to check two conditions:

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

And if we want to find the following probabilities:

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

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VashaNatasha [74]

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