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3 years ago

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Our teacher had1/2 of an eraser left.she gave that to her 2 students equally.how much of the eraser did every student take home

2 answers:

uranmaximum [27]3 years ago

7 0

1/2 an eraser is equally divided among 2 students......
1/2/2= 1/4

Thus each student receives 1/4 of the eraser

iogann1982 [59]3 years ago

5 0

Hello,

If the teacher has 1/2 of an eraser, and she divides it into 2 pieces to give 1 to each of her students, she gives (1/2) / 2 = (1/2) * (1/2) = 1/4 of the eraser to each student.

Each student takes home 1/4 of the eraser.

Hope this helps!

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the GCF of two numbers is 850 number is divisible by the other what is the smallest these two numbers can be

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If you're trying to say one number is divisible by the other, the two numbers are ...
1*850 = 850
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3 years ago

Solve the system of equations.in this form: (x, y)<br><br> -5y + 8x = -18<br><br> 5y + 2x = 58

Gre4nikov [31]

If we add the equations it looks like
-5y + 8x + 5y + 2x = -18+58
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3 years ago

Read 2 more answers

1. Given the following relation, determine the inverse relation.

gayaneshka [121]

Answer:

1. a

2. $f^{-1} (x)=\frac{x}{2} +2$

Step-by-step explanation:

1. Since there is one value of y for every value of x in (−1,0),(−4,5),(3,2),(−3,5), this relation is a function.

2. To find the inverse, interchange the variables and solve for y.

$f^{-1} (x)=\frac{x}{2} +2$

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2 years ago

A rectangle is 5cm longer that it is broad. It perimeter is 26cm. Find the length and breadth of the rectangle

Maslowich

I hope this helps you

length w+5

width w

26=2 (w+5+w)

13=2w+5

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3 0

3 years ago

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ. $e^{-\lambda.x}$

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - $e^{-\lambda.x}$

(a) <u>Probability</u> of distance at most <u>100m</u>, with λ = 0.0143:

F(100) = 1 - $e^{-0.0143.100}$

F(100) = 0.76

<u>Probability</u> of distance at most <u>200</u>:

F(200) = 1 - $e^{-0.0143.200}$

F(200) = 0.94

<u>Probability</u> of distance between <u>100 and 200</u>:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = $\frac{1}{\lambda}$

E(X) = $\frac{1}{0.0143}$

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = $\sqrt{\frac{1}{\lambda^{2}} }$

σ = $\sqrt{\frac{1}{0.0143^{2}} }$

σ = 69.93

<u>Distance exceeds the mean distance by more than 2σ</u>:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - $e^{-0.0143*209.79}$ )

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

$\int\limits^m_0 f({x}) \, dx$ = 0.5

$\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx$ = 0.5

$\lambda.\frac{e^{-\lambda.x}}{-\lambda}$ = $-e^{-\lambda.x} + e^{0}$

$1 - e^{-\lambda.m}$ = 0.5

$-e^{-\lambda.m}$ = - 0.5

ln( $e^{-0.0143.m}$ ) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0

3 years ago