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3 years ago

How would i get the letter in 42n alone?

Mathematics

2 answers:

Mice21 [21]3 years ago

5 0

You have to substitute it. For example 42n You have to either divide or multiply to get n by itself

larisa86 [58]3 years ago

4 0

42n, you can't take the "n" away. The "n" means times a number, so I I write n=2, then the expression will be 42 x 2, substitute the n with a 2

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A small tile balcony is shown . Each tile is 9 square inches .what is the area of the shaded section in square inches

DochEvi [55]

The answer to this is D. 180 square inches.

The area formula for rectangles: b*h

The base is 5 tiles.

The height is 4 tiles.

4*5=20.

Since each tile is 9 square inches, 20*9=180.

Hope this helps!

7 0

3 years ago

Solve for x what is x+5=14

ki77a [65]

Answer:

x=9

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3 years ago

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Zigmanuir [339]

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3 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

3 years ago

If DF = 170, find the value of x. Then find DE and EF.<br> DE=3x + 20<br> EF=2x + 40

Julli [10]

9514 1404 393

Answer:

DE = 86

EF = 84

Step-by-step explanation:

We assume that point E lies on segment DF, so that ...

DE + EF = DF

(3x +20) +(2x +40) = 170

5x = 110 . . . . . . . . . . . . . . . collect terms, subtract 60

x = 22 . . . . . . . . . . . . divide by 5

DE = 3×22 +20 = 66 +20 = 86

EF = 2×22 +40 = 44 +40 = 84

7 0

2 years ago