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kow [346]
3 years ago
11

Helppp its due nowwww

Mathematics
2 answers:
shepuryov [24]3 years ago
8 0

Answer: OPTION C 9x^{4\\}y^{6}

First, we need to multiply the side length with the side length, so  we will have to do the following equation

3x²y³×3x²y³

The above equstion can be written as below

⇒3×3 × x²×x² × y³×y³

Now by simple multiplication we get

⇒ 9 ×x^{4\\}×y^{6}

⇒9x^{4\\}y^{6}

Therefore the Area of the Square is 9x^{4\\}y^{6}

Sorry for  my delay and hope it helps..

Irina-Kira [14]3 years ago
6 0

Answer:

9 x^4 y^6

Step-by-step explanation:

Recall the Exponent Law:

a^m \times a^n = a^{m +n}

Solution:

3x^2y^3 \times 3x^2y^3 \\ ( 3 \times 3 )x^{2 +2} y^{3 +3} \\ 9 x^4 y^6

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The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probab
LiRa [457]

Answer:

a) 0.0082

b) 0.9987

c) 0.9192

d) 0.5000

e) 1

Step-by-step explanation:

The question is concerned with the mean of a sample.  

From the central limit theorem we have the formula:

z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }

a) z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40

The area to the left of z=2.40 is 0.9918

The area to the right of z=2.40 is 1-0.9918=0.0082

\therefore P(\bar X\:>\:1224)=0.0082

b) z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00

The area to the left of z=3.00 is 0.9987

\therefore P(\bar X\:

c) The z-value of 1200 is 0

The area to the left of 0 is 0.5

z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40

The area to the left of z=1.40 is 0.9192

The probability that the sample mean is between 1200 and 1214 is

P(1200\:

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000

e) z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65

The area to the left of z=-112.65 is 0.

The area to the right of z=-112.65 is 1-0=1

5 0
3 years ago
Pls help and no links thankyou <3
Hitman42 [59]
Do the second one hope this helps
8 0
3 years ago
Pllsss I needddd helpp
beks73 [17]
I can’t see, make it clear
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3 years ago
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