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3 years ago

Evaluate 5a+3 4b when a = b - 4and b = -2.

Mathematics

1 answer:

Svetlanka [38]3 years ago

5 0

The answer is in the photo, sorry if it isn’t correct

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Check all solutions to the equation. If there are no solutions, check "None."x2=0A. 1B. -1C.0D. 10E. 4E None

Juli2301 [7.4K]

The given equation is,

$\begin{gathered} x^2=0 \\ x=0 \end{gathered}$

Thus, option (C) is the correct solution.

8 0

1 year ago

A box contains different colored paper clips. The probability of drawing two red paper clips from the box without replacement is

ivann1987 [24]

Answer: 5/14 which is choice B

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How I got this answer:

Define the following events

A = event of picking a red paper clip on the first selection

B = event of picking a red paper clip on the second drawing

Replacement is not made.

Now onto the probabilities for each

P(A) = 2/5 = 0.4 is given to us as this is simply the probability of picking red on the first try

P(A and B) = probability of both events A and B happeing simultaneously = 1/7

P(B|A) = probability event B occurs, given event A has occured

P(B|A) = probability of selecting red on second selection, given first selection is red (no replacement)

P(B|A) = P(A and B)/P(A)

P(B|A) = (1/7) / (2/5)

P(B|A) = (1/7) * (5/2)

P(B|A) = (1*5)/(7*2)

P(B|A) = 5/14

So if event A happens, then the chances of event B happening is 5/14

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A more concrete example:

If we had 15 paperclips, and 6 of them were red, then

P(A) = (# of red)/(# total) = 6/15 = 2/5

P(B|A) = (# of red left)/(# total left) = (6-1)/(15-1) = 5/14

P(A and B) = P(A)*P(B|A) = (2/5)*(5/14) = 10/70 = 1/7

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4 years ago

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kow [346]

Answer:

Step-by-step explanation:

Pretty sure its true but im a little iffy

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3 years ago

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svlad2 [7]

Hey there, Selenawebb0418!

If you look at the table, you can see that the plant grows 2 centimeters (I'm assuming it's centimeters) every 3 days.
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Thank you for using Brainly.
See you soon!

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3 years ago

Find the derivative of<br> <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl

aev [14]

$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$

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3 years ago

Evaluate 5a+3 4b when a = b - 4and b = -2.​

Evaluate 5a+3 4b when a = b - 4and b = -2.