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2 years ago

PLEASE HELP!!!!!!!!!!!!!!

Mathematics

2 answers:

Andru [333]2 years ago

5 0

1) B

2) A

3) B

Hope this helps :)

Phoenix [80]2 years ago

4 0

1) B)0.05

2) C) 1,4,8,13

And I'm Not Sure About The Last One I Think It's C or D sorry!

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Simplify 15(10-4-2) 14 015/2 52-4110 O 105-4110

inna [77]

$\qquad\qquad\huge\underline{\boxed{\sf Answer☂}}$

let's simplify ~

$\qquad \sf \dashrightarrow \: \sqrt{5} (10 - 4 \sqrt{2} )$

$\qquad \sf \dashrightarrow \: ( \sqrt{5} \times 10) - ( \sqrt{5} \times 4\sqrt{2)}$

$\qquad \sf \dashrightarrow \: 10 \sqrt{5} - 4\sqrt{10}$

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1 year ago

What times what equals 6

Natalka [10]

Three Times Two Equals Six or Two Times Three Equals Six
Six Times One Equals Six

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2 years ago

The frequency distribution shows the lengths, in inches, of available baseball bat at a batting cage. what is the mean of the fr

ivanzaharov [21]

The mean of a frequency distribution is given by

$\bar{x}= \frac{\Sigma fx}{\Sigma f} \\ \\ = \frac{(27\times2)+(28\times5)+(31\times3)+(33\times1)+(34\times4)}{2+5+3+1+4} \\ \\ = \frac{54+140+93+33+136}{15} = \frac{456}{15} =30.4$

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3 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

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).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago

Which numbers in set ={,,,,,,,,,}eh equals left brace 1 comma 2 comma 3 comma 4 comma 5 comma 6 comma 7 comma 8 comma 9 comma 10

viva [34]

Answer:

D={3,9}

The numbers are 3 and 9

Step-by-step explanation:

The set A

={,,,,,,,,,}

Let B be the sub set of A containing odd numbers

B={1,3,5,7,9}

Let C be the sub set of A containing multiply of 3

C= {3,6,9}

Now let D be the be the sub set of A containing both odd numbers and multiples of 3

D={3,9}

5 0

2 years ago