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3 years ago

Please help!! I really need this for geometry so I won't fail the class.

Mathematics

2 answers:

Sindrei [870]3 years ago

7 0

2. c. 3. c. i hope this helps

wel3 years ago

5 0

2. C. 9 miles
3. B. 56 inches
To find the perimeter you have to add the lengths of each side.
(12+16)+(12+16)
28+28
=56

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Bethany and Aidan each have candy bars that are the same size and shape. Bethany's candy bar is marked so that it can be broken

vivado [14]

Answer:

Step-by-step explanation:

This problem is most likely about what percentage of the candy does each individual have left. Since the phrase is explaining that each ones candy bar is the same size and can be broken into equally divided pieces then it is most likely about figuring out which one of the individual's has more candy left over when compared to the other's candy. This is possible since both candies are the same size but more information would be needed.

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Which of these professionals most directly uses geometry?

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4 years ago

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3 years ago

Suppose that a college determines the following distribution for X = number of courses taken by a full-time student this semeste

lidiya [134]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$ In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$ And the variance is given by:

$Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924$

And the deviation would be:

$Sd(X) =\sqrt{0.8924} =0.9447$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

For this case we have the following distribution given:

X 3 4 5 6

P(X) 0.07 0.4 0.25 0.28

We can calculate the mean with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$

In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$

And the variance is given by:

$Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924$

And the deviation would be:

$Sd(X) =\sqrt{0.8924} =0.9447$

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4 years ago

Sean got m marks for his math test.Emma got 15 more marks than Sean Emma score is 97

Dafna1 [17]

97-15=82 so he got 82

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3 years ago