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Rina8888 [55]
3 years ago
9

PLSSSSSS!!!! HELP ME I WILL MARK YOU!!!!!!!!!!

Mathematics
2 answers:
Olegator [25]3 years ago
8 0

Answer:

2.25? i think thats right........

joja [24]3 years ago
5 0

Answer:

2.25

Step-by-step explanation: Replace the z with the value given (-18) and then multiply that by 1.1 you can then add it to 21.3 and then do the exponent

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2y+8×=2 identify the table that would correctly graph this equation
lys-0071 [83]

We have

2y+8x=2

In order to obtain easily the table, we need to clear y

\begin{gathered} 2y=-8x+2 \\ y=\frac{-8x+2}{2} \\ y=-4x+1 \end{gathered}

then we evaluate for values of x

if x=0

y=-4(0)+1=1

y=1

if x=1

y=-4(1)+1=-3

y=3

if x=2

y=-4(2)+1=-7

y=-7

if x=3

y=-4(3)+1=-11

y=-11

So the table for the given equation is

x y

0 1

1 -3

2 -7

3 -11

3 0
1 year ago
What is the area of this regular pentagon that has been divided into five congruent triangles?
Len [333]

Answer:

120 cm^2.

Step-by-step explanation:

The area of one triangle = 1/2 * base * height

= 1/2 * 8 * 6

= 1/2 * 48

= 24 cm^2

So area of the pentagon = 5 * 24

= 120 cm^2.

7 0
3 years ago
Pls help on #10 will give brainliest
Free_Kalibri [48]
The answer is A hope this helps
4 0
3 years ago
The polygons are similar. Find the missing side length.
zloy xaker [14]

Answer:

44

Step-by-step explanation:

Since triangles are similar, so their corresponding sides would be in proportion.

Therefore,

\frac{24}{32}  =  \frac{33}{?}  \\  \\  \\ ? =  \frac{32 \times 33}{24}  \\  \\ ? =  \frac{4\times 33}{3} \\  \\ ? =  4\times 11 \\  \\  ? =  44 \: units

6 0
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
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