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Elden [556K]
2 years ago
11

PLEASE HELP ME I don’t know how to solve this one please give me a good explanation

Mathematics
2 answers:
Luba_88 [7]2 years ago
4 0

Answer:

12

Step-by-step explanation:

assuming you wish to evaluate the expression.

substitute x = 5 into the expression

\frac{2(5+13)}{3}

= \frac{2(18)}{3}

= \frac{36}{3}

= 12

forsale [732]2 years ago
3 0
You have to use the distributive property, therefore you get 2x+13/3.

The next step is to do 2 x 5 because they gave us a number for the variable x. Then you have to do 2 x 13 which gives us 26.

10+26=36 and 36/3=12, hope I helped, Have a nice day! :)
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In the diagram below BD is parallel to XY what is the value of Y
LiRa [457]

Given:

BD is parallel to XY.

One of the angle is 85°

To find:

The value of y.

Solution:

Parallel lines BD and XY cut by a transversal line.

Angle y and 85° are alternate interior angles.

Alternate interior angle theorem:

If two parallel lines are cut by a transversal line, then the pairs of alternate interior angles are congruent.

⇒ y = 85°

The value of y is 85.

4 0
3 years ago
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Two consecutive integers are such that 7 times the biggest and then added to the smallest equals
Arturiano [62]

Answer:

61,62

Step-by-step explanation:

Let the smallest  number be x

Hence, the biggest number = x+1 [as it consecutive]

Hence,

7(x+1)+x=495

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8x=495-7

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7 0
3 years ago
GUYS PLEASE HELP ME PLEASE HELP PLEASE PLEASE!!!!
snow_tiger [21]

Answer:

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Step-by-step explanation:

3.052 * 6.193

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3 0
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:

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

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