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KATRIN_1 [288]
3 years ago
11

Sara brought a piece of ribbion. The length of the ribbion is given in tenths. Write the length as two other equivalent fraction

s. Solve this problem any way you choose. (The meter is 6/10)
Mathematics
2 answers:
hram777 [196]3 years ago
8 0

Divide both 6 and 10 by 2 to get 3/5

Multiply both 6 and 10 by 2 to get 12/20

luda_lava [24]3 years ago
6 0

Good morning,

Answer:

3/5 , 60/100

Step-by-step explanation:

To get an equivalent fraction to 6/10 just multiply or divide both numerator and denominator by the same number

e.g 1 : 6/10 = (6×2)/10×2) = 3/5

e.g 2 : 6/10 = (6×10)/(10×10) = 60/100.

:)

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Graph the line by locating any two ordered pairs that satisfy the equation. Round to the nearest thousandth, if necessary.
solmaris [256]

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To do this we enter a value of "x" and then find "y"

Let's start with x = 2

y = 2 (2)

y = 4

Then the first ordered pair that satisfies the equation is dot (2,4)

Now, for x = 0

y = 2(0)

y = 0

Then the second ordered pair that satisfies the equation is the point (0,0).

With these two pairs we draw the graph of the equation, which is shown in the attached image.

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2 years ago
An experiment consist of rolling two fair number cubes. The diagram shows the sample space of all equally likely outcomes. Find
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3 years ago
Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha
const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

P(1.83\leq X\leq  a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06

Thus, the value of <em>a</em> is 12.06.

7 0
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Natasha_Volkova [10]

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x has a value of 1. Use this value to find the value of y.

\begin{gathered} y=5x-4 \\ y=5\cdot1-4 \\ y=5-4 \\ y=1 \end{gathered}

y has a value of 1.

x=1

y=1

8 0
1 year ago
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