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Svetradugi [14.3K]
2 years ago
11

Prove that parallelogram circumscribing a circle is a rhombus​

Mathematics
1 answer:
Romashka [77]2 years ago
6 0

Answer:

Answer:

Step-by-step explanation:

I hope it's helpful!

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Without dividing or multiplying, explain why 0.8 divided by 0.7 is greater
Art [367]

Answer:

It is greater because when you divide decimals you move the decimal which would automatically make the product less than the quotient.

Step-by-step explanation:

3 0
3 years ago
if their is a family of 9 and their are 4 boys and 5 girls what is the percentage percentage of both male and female
gregori [183]
Boys 4/9 = 44.44% Girls 5/9 = 55.55%

Fraction to percentage: Divide the top by the bottom then multiply the result by 100
6 0
3 years ago
4.4.3 Quiz: Independence
lbvjy [14]

Answer:

\frac{2}{15}

Step-by-step explanation:

<h2><u>Lesson : Probability</u></h2><h2><u></u></h2>
  • We have a bag with 4 white chips and 6 black chips
  • Question : What is the probability of randomly choosing a white chip, not replacing it, and then randomly choosing another white chip?

4 white and 6 black

4 + 6 = 10

Choosing the first white chip : \frac{2}{5}

Since we took one out , now

3 white chips , 6 black chips

3 + 6 = 9

Choosing another white chip : \frac{1}{3}

\frac{2}{5} * \frac{1}{3}

\frac{2}{15}

<u></u>

<u></u>

7 0
3 years ago
Factor -8m^3n^3+2m^5+2m^3
Arada [10]

−

2

m

3

 out of  

−

8

m

3

n

3

+

2

m

5

+

2

m

3

.

−

2

m

3

(

4

n

3

−

m

2

−

1

)

7 0
3 years ago
Suppose a certain type of fertilizer has an expected yield per acre of mu 1 with variance sigma 2, whereas the expected yield fo
mart [117]

Answer:

See the proof below.

Step-by-step explanation:

For this case we just need to apply properties of expected value. We know that the estimator is given by:

S^2_p= \frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2}

And we want to proof that E(S^2_p)= \sigma^2

So we can begin with this:

E(S^2_p)= E(\frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2})

And we can distribute the expected value into the temrs like this:

E(S^2_p)= \frac{(n_1 -1) E(S^2_1) +(n_2 -1) E(S^2_2)}{n_1 +n_2 -2}

And we know that the expected value for the estimator of the variance s is \sigma, or in other way E(s) = \sigma so if we apply this property here we have:

E(S^2_p)= \frac{(n_1 -1 )\sigma^2_1 +(n_2 -1) \sigma^2_2}{n_1 +n_2 -2}

And we know that \sigma^2_1 = \sigma^2_2 = \sigma^2 so using this we can take common factor like this:

E(S^2_p)= \frac{(n_1 -1) +(n_2 -1)}{n_1 +n_2 -2} \sigma^2 =\sigma^2

And then we see that the pooled variance is an unbiased estimator for the population variance when we have two population with the same variance.

8 0
3 years ago
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