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Alekssandra [29.7K]
3 years ago
10

Select the fraction that is equal to 2/4 1/5 12/7 1/2

Mathematics
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

1/2

hope this helps

have a good day :)

Step-by-step explanation:

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A line passes through (-1, 5) and (1, 3)
Luba_88 [7]

Answer:

y=-x+4

Step-by-step explanation:

3-5/1-(-1) = -1

(y-(-1)) = -1(x-5)

y+1 = -x+5

y= -x +5-1

y= -x + 4

6 0
3 years ago
Use the distributive property to factor the expression below. 5x2 + 25
ollegr [7]
First you need to find the highest common factor of 5x² and 25, which is 5 because there is no x term in 25. This means that 5 goes on the outside of the brackets, like 5 (______). To get 5x² from 5, you need to multiply by x², which is your first term in the brackets. To get 25 from 5, you need to multiply by 5, which would be your second term, so your final answer is 5(x² + 5). I hope this helps!
6 0
3 years ago
5/2x+7.75=y in standard form
Naddika [18.5K]
Subtract 7.75 by both sides then. subtract y both sides. you need x and y to be on one side. 5/2x-y=-7.75
3 0
4 years ago
Hiii i have a question on math:)
vlabodo [156]
  • <em>Answer:</em>

<em>PQR = 150°</em>

  • <em>Step-by-step explanation:</em>

<em>Hi there ! </em>

<em>PQR = PQS + SQR</em>

<em>12x - 6 = 16 + 9x + 17</em>

<em>12x - 9x = 16 + 17 + 6</em>

<em>3x = 39</em>

<em>x = 13</em>

<em>PQR = (12×13 - 6)° = (156 - 6)° = 150°</em>

<em>Good luck !</em>

4 0
3 years ago
A company is studying the number of monthly absences among its 125 employees. The following probability distribution shows the l
Angelina_Jolie [31]

Answer:

(1)

E(x) = 0.72

Var(x) = 1.1616

\sigma = 1.078

(2)

E(x) = 12

Var(x) = 1.3

\sigma = 1.14

Step-by-step explanation:

Solving (1):

\begin{array}{ccccccc}{Days} & {0} & {1} & {2} & {3} & {4}& {5} \ \\ {Probability} & {0.60} & {0.20} & {0.12} & {0.04} & {0.04} & {0.00} \ \end{array}

(a): Mean

This is calculated as:

E(x) = \sum x * P(x)

So, we have:

E(x) = 0 * 0.60 + 1 * 0.20 + 2 * 0.12 + 3 * 0.04 + 4 * 0.04 + 5 * 0.00

E(x) = 0.72

Solving (b): The variance

This is calculated as:

Var(x) = E(x^2) - (E(x))^2

Where:

E(x) = 0.72

and E(x^2) = \sum x^2 * P(x)

So, we have:

E(x^2) = 0^2 * 0.60 + 1^2 * 0.20 + 2^2 * 0.12 + 3^2 * 0.04 + 4^2 * 0.04 + 5^2 * 0.00

E(x^2) = 1.68

So, we have:

Var(x) = E(x^2) - (E(x))^2

Var(x) = 1.68 - 0.72^2

Var(x) = 1.1616

Solving (c): The standard deviation.

This is calculated as:

\sigma = \sqrt{Var(x)}

\sigma = \sqrt{1.1616}

\sigma = 1.078

Solving (2):

\begin{array}{ccccccc}{x} & {10} & {11} & {12} & {13} & {14}& { } \ \\ {P(x)} & {0.1} & {0.25} & {0.3} & {0.25} & {0.1} & { } \ \end{array}

(a): Mean

This is calculated as:

E(x) = \sum x * P(x)

So, we have:

E(x) = 10 * 0.10 + 11 * 0.25 + 12 * 0.3 + 13 * 0.25 + 14 * 0.1

E(x) = 12

Solving (b): The variance

This is calculated as:

Var(x) = E(x^2) - (E(x))^2

Where:

E(x) = 12

and E(x^2) = \sum x^2 * P(x)

So, we have:

E(x^2) = 10^2 * 0.10 + 11^2 * 0.25 + 12^2 * 0.3 + 13^2 * 0.25 + 14^2 * 0.1

E(x^2) = 145.3

So, we have:

Var(x) = E(x^2) - (E(x))^2

Var(x) = 145.3 - 12^2

Var(x) = 1.3

Solving (c): The standard deviation.

This is calculated as:

\sigma = \sqrt{Var(x)}

\sigma = \sqrt{1.3}

\sigma = 1.14

8 0
3 years ago
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