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3 years ago

Write an equation in slope intercept form given m: m=-3/5 and b=-3

Mathematics

1 answer:

brilliants [131]3 years ago

4 0

Step-by-step explanation:

Y=kx+b

If m=1/k

Y=-5/3x+b

I don't know what your m means

So I can only write like this.

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Write the given sentence as an equation.

Leni [432]

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4 years ago

How to round to the nearest degree

Anika [276]

Answer:

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3 years ago

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3 years ago

Write an equation of the line passing through each of the following pairs of points. b (5, 7), (−6, −3)

sdas [7]

Answer:

y = 10/11 x + 2.45

Step-by-step explanation:

Equation of the line

First find the gradient

formula = $\frac{y2-y1}{x2-x1}$

= $\frac{-3-7}{-6-5}$

= $\frac{-10}{-11}$

= 10/11

m is the gradient

equation of a line

y- y1 = m(x-x1)

y-7 = 10/11 (x-5)

y-7 = $\frac{10}{11}x$ - $\frac{50}{11}$

y = 10/11 x + 2.45

NB i rounded of -50/11 +7 = 2.4545..... to 2.45

3 0

3 years ago

Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte

sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so $n = 11$

Each player with a 5% probability of injury, so $p = 0.05$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688$

$P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293$

$P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848$

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so $n = 2$

Each team has a 0.9848 probability of showing up to play, so $p = 0.9848$ .

This probability is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299$

2.99% probability that the cost will be paid by only one team

7 0

4 years ago