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1 year ago

¿Cuál es la ecuación de la recta con pendiente -2 y el punto (4, 10)?

Mathematics

1 answer:

Sophie [7]1 year ago

4 0

The general slope-intercept form equation is:

$\begin{gathered} y=mx+b \\ \text{Wher m is the slope and b is the y-intercept value} \end{gathered}$

In this case, m=-2 and with the point (4, 10) we can find the value of b:

$\begin{gathered} We\text{ evalute the equation in the point (x, y) = (4 , 10):} \\ 10=-2\cdot4+b \\ 10=-8+b \\ b=10+8=18 \end{gathered}$

We found that b = 18, so:

$y=-2x+18$

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F(x)=-3/5x+k and if f(30)=-11 what is the value of k

Kruka [31]

Answer:

k=20.

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Find the sum simplify answer 6 2/3 + 5 1/2=

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3 years ago

Read 2 more answers

Answer the following equations:

Andrew [12]

Step-by-step explanation:

3(t+5) = 9

3t+15 = 9

3t = -6

t = -2

2(f-7) = -10

2f - 14 = -10

2f = 4

f = 2

-(c - 9) = 4

-c + 9 = 4

-c = -5

c = 5

-6(2t + 8) = -84

-12t - 48 = -84

-12t = -36

12t = 36

t = 3

-10 (s + 2) = -57

-10s - 20 = -57

-10s = -37

s = 3.7

7(3w + 8)/3 = -9

By cross multiplication,

7(3w + 8) = -27

21w + 56 = -27

21w = -83

w = -3.95

35/5 = (F - 32)/9

7 = (F - 32)/9

By cross multiplication,

63 = F - 32

63 + 32 = F

95 = F

Hope it helps.

7 0

2 years ago

The dingram shows a spinner innde up of a picce of card in the shape of a regular pentagon, with a toothpick pushed through its

Anastaziya [24]

Answer:

a) $P(X \leq 3) = 0.99328$

b) 0.6957

Step-by-step explanation:

Let X represent the number of 4's when n = 5 independent spins

each has a probability of 0.2 (i.e p = 0.2)

This notation is represented as:

X $\approx$ Binomial (n = 5, p = 0.2)

Probability of $x$ number of 4's is:

$P(X=x)= (\left \ n \atop x \right) p^x (1-p)^{(n-x)}$

here; $(\left \ n \atop x \right)$ is the combinatorial expression

$(\left \ n \atop x \right)$ = $\frac{n!}{x!(n-x)!}$

$P(X \leq3), n =5 , p = 0.2$

$P(X \leq3) = 1-P(X > 3)$

So; let's first find:

$P(X > 3)$

$= P(3$

$P(X = 4) =( \left \ {{5} \atop {4}} \right. ) (0.2)^4 (1-0.2)^1 \\ \\ P(X = 4) = 5(0.2)^4(0.8)^1 \\ \\ P(X = 4) = 0.0064$

$P(X = 5) =( \left \ {{5} \atop {5}} \right. ) (0.2)^5 (1-0.2)^0 \\ \\ P(X = 5) = 5(0.2)^5(0.8)^0 \\ \\ P(X = 5) = 0.00032$

$P (X=4)+P(X = 5 ) \\ \\ = 0.0064 + 0.00032 = 0.006720 \\ \\ \approx 0.007$

$P(X > 3 ) = 0.00672 \\ \\ P(X \leq 3) = 1- P(X > = 3 ) \\ \\ =1 - 0.00672 \\ \\ = 0.99328$

$P(X \leq 3) = 0.99328$

Given that:

The ratio of boys to girls at birth in Singapore is quite high at 1.09:1

What proportion of Singapore families with exactly 6 children will have at least 3 boys?

Probability of having a boy = $\frac{1.09}{1+1.09}$ = 0.5215

Binomial Problem with n = 6

P(3<= x <=6) = 1 - P(0<= x <=2)

= 1 - binomial (6,0.5215,2)

= 0.6957

5 0

3 years ago