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3 years ago

Plz help me on this question

Mathematics

1 answer:

Deffense [45]3 years ago

4 0

Let number $c=\sqrt{45}.$

Square this number:

$c^2=45.$

Since

$36$

you can state that

$\sqrt{36}$

Also $\sqrt{45}\approx 6.708$ and $6.5$ , then point C is the best approximation of $\sqrt{45}.$

Answer: option C.

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2 years ago

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A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

bulgar [2K]

Answer:

a) 20.61%

b) 21.82%

c) 42.36%

d) 4 withdrawals

Step-by-step explanation:

This situation can be modeled with a binomial distribution, where p = probability of “success” (completing the course) equals 80% = 0.8 and the probability of “failure” (withdrawing) equals 0.2.

So, the probability of exactly k withdrawals in 20 cases is given by

$\large P(20;k)=\binom{20}{k}(0.2)^k(0.8)^{20-k}$

We are looking for

P(0;20)+P(0;1)+P(0;2) =

$\large \binom{20}{0}(0.2)^0(0.8)^{20}+\binom{20}{1}(0.2)^1(0.8)^{19}+\binom{20}{2}(0.2)^2(0.8)^{18}=$

0.0115292150460685 + 0.0576460752303424 + 0.136909428672063 = 0.206084718948474≅ 0.2061 or 20.61%

Here we want P(20;4)

$\large P(20;4)=\binom{20}{4}(0.2)^4(0.8)^{16}=0.218199402\approx 0.2182=21.82\%$

Here we need

$\large \sum_{k=4}^{20}P(20;k)=1-\sum_{k=1}^{3}P(20;k)$

But we already have P(0;20)+P(0;1)+P(0;2) =0.2061 and

$\large \sum_{k=1}^{3}P(20;k)=0.2061+P(20;3)=0.2061+0.205364 \approx 0.4236=42.36\%$

For a binomial distribution the <em>expectance </em>of “succeses” in n trials is np where p is the probability of “succes”, and the expectance of “failures” is nq, so the expectance for withdrawals in 20 students is 20*0.2 = <em>4 withdrawals.</em>

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3 years ago

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Answer:

Step-by-step explanation:

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3 years ago

The interior angle of a regular polygon is five times its corresponding exterior angle. Find the

tiny-mole [99]

Answer:

Step-by-step explanation:

Regular polygon is one in which each angle and sides is congruent.

Sum of all the exterior angles of any polygon is 360.

Sum of all the interior angles of any polygon is (2n-4)*90

where is n is the no. of sides of polygon.

Let n be the no. of sides of polygon required

Therefore,

Sum of all the interior angles of polygon = (2n-4)*90

since no of sides of polygon is n

therefore, value of each interior angles of the polygon = (2n-4)*90/n (A)

Sum of all the exterior angles of any polygon = 360.

value of each exterior angles of the polygon = 360/n (B)

Given that

The interior angle of a regular polygon is five times its corresponding exterior angle (c)

using the statement A, B and C

(2n-4)*90/n = 5*360/n

1/n is common on both side, hence it gets cancelled.

(2n-4)*90 = 5*360

=> (2n-4) = 5*360 /90 = 20

=> 2n = 20+4 = 24

=> n = 24/2 = 12.

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