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expeople1 [14]
3 years ago
11

I need help with scale drawings ​

Mathematics
1 answer:
Bogdan [553]3 years ago
8 0
Umm ok where is the picture
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What is the equation of this line in slope-intercept form?
KatRina [158]

Answer: B

Step-by-step explanation:

The slope-intercept is usually written as y = mx + b form, where m is the slope, and b is the y-intercept. The y-intercept is the y value when x is equal to 0.

We can first eliminate C, because the y-intercept of the graph is 1.

To solve for slope, which is the m, choose two points, (-1, 5) and (1, -3).

Slope = (y2 - y1) / (x2 - x1)  

           = (-3 - 5) / (1 - (-1))

          = (-3 - 5) / (1 + 1)

          = -8/2

          = -4

Plug in the slope and y-intercept, we get y = -4x + 1

       

5 0
3 years ago
Evaluate -x - y when x = -12 and y = 2
xz_007 [3.2K]

Answer:

-14

Step-by-step explanation:

subtract

8 0
3 years ago
Read 2 more answers
Solve this application problem using a system of equations: The Bright College footballteam scored 80 times last season, some on
Goryan [66]

Answer:

\large \boxed{\sf \ \ 69 \ \textbf{touchdowns } 11 \ \textbf{field goals } \ }

Step-by-step explanation:

Hello,

Touchdown is 7 points, field goals is 3 points

Let's note a the number of touchdowns and b the number of field goals, we can write

   (1)    a + b = 80

   (2)    7a + 3b = 516

(2) - 3*(1) gives

7a + 3b - 3(a + b) = 516 - 3*80

<=> 7a + 3b -3a - 3b = 516 - 240 = 276

<=> 4a = 276

<=> a=\dfrac{276}{4}=69

And then from (1) b = 80 - 69 = 11

Hope this helps

4 0
3 years ago
The sum of two exterior angles of an isosceles triangle is equal to 240°. What are the measures of the interior angles? THERE AR
Alla [95]
Dang that’s crazy, send me some pics and you got a deal
8 0
3 years ago
The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that
salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

Part a

Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0
4 years ago
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