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2 years ago

Which expression is an irrational number?

Mathematics

1 answer:

Alex2 years ago

5 0

✔8 Isn't a perfect square. Irrational numbers are numbers that cannot be represented as a ratio for example Pi, or non-perfect squares like ✔2, ✔3, ✔5, ✔8.

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Name the transformations of the next equation y=3(x+6) - 5

klemol [59]

Answer:

vertical stretch

Step-by-step explanation:

8 0

2 years ago

Please help ASAP, No links or report, Tysm!

Margarita [4]

Answer:

<u>96 ft</u> is the perimeter of the figure.

Step-by-step explanation:

9 + 9 + 9 + 9 + 9 + 9 + 12 + 30 = 96

Have a nice day!

4 0

2 years ago

Read 2 more answers

How to tell if an ordered pair is a solution to a function

Evgen [1.6K]

A) yes
B) yes
C) no

For each of these, substitute the value of x in the ordered pair into x in the function.

For A, x = -5; -5<2, so the piece of the function we want is f(x) = 3. In our ordered pair, y=f(x)=3, so yes, it is a solution.

For B, x = 2; 2≤2<6, so the piece of the function we want is f(x) = -x+1. In our ordered pair, y=f(x)=-1; -2+1=-1, so yes, it is a solution.

For C, x = 8; 8≥6, so the piece of the function we want is f(x) = x. In our ordered pair, y=f(x)=-7; -7≠8, so no, it is not a solution.

5 0

3 years ago

For function of g,g(0)=1 and g(1)=5 assuming g is a linear function what is g(3)<br>

lawyer [7]

Answer:

$g(3) = 13$

Step-by-step explanation:

Given

$g(0) = 1$

$g(1) = 5$

Required

Find g(3)

Since g(x) is a linear function, then:

$g(x) = mx + c$

For: $g(0) = 1$

$g(x) = mx + c$

$g(0) = m*0 + c$

$1 = 0 + c$

$1 = c$

$c = 1$

For: $g(1) = 5$

$g(x) = mx + c$

$g(1) = m * 1 + c$

$5 = m * 1 + c$

$5 = m + c$

Substitute $c = 1$

$5 = m + 1$

$m = 5 - 1$

$m =4$

So, the function is:

$g(x) = mx + c$

$g(x) = 4x + 1$

Calculate g(3)

$g(3) = 4*3 + 1$

$g(3) = 12 + 1$

$g(3) = 13$

3 0

3 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