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3 years ago

Are the perimeter and side length of squares proportional? Explain.

Mathematics

1 answer:

Elan Coil [88]3 years ago

5 0

yes because if you look you can see a pattern of 4 if we divide 2 AND 8 WE GET 4 AND SO.

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I need 14 and 15 thanks

Bad White [126]

14) The answer is a) y = 2x. 2 is the slope and the line does not cross over the Y axis in this picture so there is not a Y-intercept you can see.
15) You pretty much just estimate. On the graph when you find 50 minutes and go up its around 140. 50 x 2 is 100 minutes. so just multiply 140 by 2 and its 280 so just round to the closest one.

8 0

3 years ago

Please help and show work!!

serious [3.7K]

17.

x = -2 is not a solution of -1 < x < 5 because -2 < -1 (-1 < -2 < 5 FALSE).

18.

m = 5 is a solution of 5 ≤ m because 5 ≤ 5 ( 5 ≤ m → m ≥ 5 greater than 5 or equal 5, 5 is equal 5)

19.

k = 10 is not solution of 2k - 3 < 1 because:

put the value of k to the inequality:

2(10) - 3 < 1

20 - 3 < 1

17 < 1 FALSE

5 0

2 years ago

Can you tern this into a equation

Aleks [24]

Answer:

n + p x 3 = t

Step-by-step explanation:

n= Navita's money

p= Pawan's money

t= total money

8 0

2 years ago

the area of a parallelogram os 150 square meters the height of the parallelogram is 15 meters what is the length of the parallel

lys-0071 [83]

The length is 10 meters.

150 divided by 15 is 10.

Hope this helps!

3 0

3 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

3 years ago