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2 years ago

How do i draw two number lines that show 0.20 and 1/5 are equivalent?

1 answer:

6 0

The frist line may be one with the numbers 0, 1 and 2 and 10 divisions (marks at equal distance) between each integer. Each division will be equal to 0.1 units and then you can mark the second division from the zero point to the right as the 0.20 mark.

The other line must have the same integers, 0 , 1 and 2 placed in identical form as the first line. Then
   - draw an inclined straight line since the point zero,
   - mark 5 points in the inclined lined equally spaced over the line.
   - draw a sttraight line from the 5th point to the point with the mar 1 over the base number line.
   - draw a parallel line to the previous one passing trhough the second point of the inclined line and mark the point where this parallel touchs the base number line. This point shall be at the same distance from zero than the 0.2 mark was in the first number line, meaning that 0.2 and 1/5 are equivalent.

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Read 2 more answers

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Step-by-step explanation:

<h2><em><u>You can solve this using the binomial probability formula.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>

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3 years ago

Line 1 passes through the points -2,7 and 3,3. Line 2 passes through the points 5,1 and 9,1. Find the slope of each line. Which

RSB [31]

Answer:

The slope of the given line 1 is -4/5.

The slope of the given line 2 is 0.

The line 2 has a greater slope then line 1.

Step-by-step explanation:

The slope of a line with points (x1,y1) and (x2,y2) is given as:

$m = \frac{y_2- y_1}{x_2-x_1}$

Now, Here the, points of the line 1 are A (-2,7) and B (3,3)

So, here, the slope of line AB is $m = \frac{3 -7}{3-(-2)} = \frac{-4}{3 +2} = \frac{-4}{5}$

Hence, the slope of the given line 1 is -4/5.

Now, Here the, points of the line 2 are P (5,1) and Q (9,1)

So, here, the slope of line PQ is $m = \frac{1-1}{9-5} = \frac{0}{4} = 0$

Hence, the slope of the given line 2 is 0.

Now, as we can see 0 > ( -4 / 5)

So, the line 2 has a greater slope then line 1.

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2 years ago