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2 years ago

All Questions Are Above

Mathematics

1 answer:

kompoz [17]2 years ago

5 0

Answer:

2. 5.25

3. 3

4. 81.68

Step-by-step explanation:

Hope this helps and sorry if i'm wrong

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It is answer choice A. (-1, 2.5)

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Based on the following system of equations, determine the number of solutions.

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In ∆ABC, point D is the centroid. If (AE) ̅=15, (BF) ̅=27, and (CD) ̅=8, find the following lengths

Anna71 [15]

Answer:

5,18,12 cms are the answer.

Step-by-step explanation:

Given is a triangle ABC. Point D is the centroid.

E,F and G are midpoints of CB, BA and AC respectively.

AE, BF and CG are medians of the triangle.

We know that centroid divides the median in the ratio 2:1

Using this we find that AD:DE = 2:1

Or AD+DE:DE = (2+1):1

AE:DE =3:1

15:DE = 3:1 . Hence DE =5 cm.

On the similar grounds we find that DF = 1/3 BF = 9

Hence BD = DF-BF = 27-9 =18 cm

and also

CG = 3/2 times CD = 12 cm.

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3 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

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Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$

Part C

$\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971$

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musickatia [10]

0.6cm or 6mm if you take 5*12% you can get the answer pretty easy :)

I would appriciate Brainliest!

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