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2 years ago

Hellpppppp please I will give brainliest

Mathematics

2 answers:

liberstina [14]2 years ago

8 0

Answer:

y = 5/4x - 1

Step-by-step explanation:

if you look at the graph you will probably be able to tell

im not trying to be me if you think

Vedmedyk [2.9K]2 years ago

4 0

Answer:

ITS THE 2ND ONE

Step-by-step explanation:

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Simplify. 14+{−2+3[1+3(−6−2)]} show how to solve as well with answer

frez [133]

It’s -57 thank me later

3 0

3 years ago

Read 2 more answers

The longest side of the right triangle is 2 feet more than the middle side and the middle side is 1 foot less than twice the sho

Ket [755]

A right triangle's longest side is the hypotenuse
let x=longest, y=middle, and z=shortest
x=y+2
y=2z-1
therefore x=(2z-1)+2=2z+1
find z
z^2+y^2=x^2 by Pythagorean theorem
plug in x and y in terms of z
z^2+(2z-1)^2=(2z+1)^2
z^2+4z^2-4z+1=4z^2+4z+1
subtract the right-hand side's value from the left-hand side's
z^2-8z=0
z(z-8)=0
z=0, 8
z cannot be zero as the sides must have some value to it.
Therefore the shortest side is equal to 8

3 0

3 years ago

Find 5 solutions to the question. Y=-1.5x+2

Black_prince [1.1K]

Answer:

☆< 《HOPE IT WILL HELP YOU》>☆

Step-by-step explanation:

Y=1.5x+2

y=1.5 (1) +2

y=1.5+2

y=3.5

y=1.5(2)+2

y=3+2

y=5

y=1.5 (3)+2

y=4.5+2

y=6.5

y=1.5 (4)+2

y=6+2

y=8

y=1.5 (5)+2

y=7.5+2

y=9.5

<h2>please mark my ans as BRAIN LIST</h2>

3 0

3 years ago

Solve for x pls help me

Usimov [2.4K]

Answer:

Step-by-step explanation:

circumference of total circle = 16 pi

16 pi / 4 pi = 4

1 / 4 =25%

1/4 * 360 = 90

4 0

2 years ago

You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob

Ilia_Sergeevich [38]

Answer:

a) No

b) 42%

c) 8%

d) X 0 1 2

P(X) 42% 50% 8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c) P(win first game) = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X 0 1 2

P(X) 42% 50% 8%

P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value $\mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66$

f) Variance $\sigma^2=\Sigma(x-\mu^2)p(x)= (0-0.66)^2*0.42+ (1-0.66)^2*0.5+ (2-0.66)^2*0.08=0.3844$

Standard deviation $\sigma=\sqrt{variance} = \sqrt{0.3844}=0.62$

8 0

3 years ago

Hellpppppp please I will give brainliest​

Hellpppppp please I will give brainliest