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2 years ago

HELP ITS MY HW PLZ o0o

Mathematics

2 answers:

tamaranim1 [39]2 years ago

8 0

Q2 = a -14, Q3 = f+4, Q4 = c + 25, Q5 = h + 8, Q6 = m + 43

murzikaleks [220]2 years ago

7 0

Can u move the camera tot he left more, i cant see the boxes and questions

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Julli [10]

The answer I think is C

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2 years ago

If a segment has an endpoint at (3, 2) and the midpoint at (-1, 2), what are the coordinates of the other endpoint?

Paraphin [41]

The other endpoint would have to be (-5,2), going off the definition of a midpoint. Since 3 is for units to the right of -1, that would mean that the other end point would have to be 4 units to the left, which is -5.

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2 years ago

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Let X be the number of Heads in 10 fair coin tosses. (a) Find the conditional PMF of X, given that the first two tosses both lan

inn [45]

Answer:

Follows are the solution to the given point:

Step-by-step explanation:

For option A:

In the first point let z be the number of heads which is available on the first two trails of tosses so, the equation is:

$P(X=k | z=2 ) = \begin{pmatrix} 10-2\\ k-2\end{pmatrix} (\frac{1}{2})^{k-2} (\frac{1}{2})^{10-k}$

$= \begin{pmatrix} 8\\k-2\end{pmatrix} (\frac{1}{2})^{k-2} (\frac{1}{2})^{10-k} \ \ \ \ \ \ \ \ \ \ \\\\$

$k= 2,3..........10$ For option B:

$P(X=k | X \geq 2 ) = \sum^{10}_{i=2} \begin{pmatrix} 10-i\\ k-i\end{pmatrix} (\frac{1}{2})^{k-i} (\frac{1}{2})^{10-k}$

$k= 2,3, 4.........10$

8 0

2 years ago

Please help asap thx a bunch

Aloiza [94]

it would be high I think

3 0

2 years ago

Find the arc length and area of a sector with radius R =60 inches and central angle equals 30° round your answer to the nearest

Black_prince [1.1K]

Answer:

(a)31.42 inches

(b)942.48 Square Inches

Step-by-step explanation:

Given a sector of a circle with the following dimensions:

Radius of the circle =60 inches

Central Angle of the sector =30°

(a)Arc Length

Arc Length $=\dfrac{\theta}{360^\circ}X2\pi r$

$=\dfrac{30}{360^\circ}X2*60*\pi\\\\=10\pi\\\\=31.42$ Inches (correct to the nearest hundredth)$

(b)Area of the sector

Area of the sector $=\dfrac{\theta}{360^\circ}X\pi r^2$

$=\dfrac{30}{360^\circ}X\pi*60^2\\\\=300\pi\\\\=942.48$ Square Inches (correct to the nearest hundredth)$

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2 years ago

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