This question has 2 parts.

Which step in the solution contains the first error ?? Please helpp

kumpel [21]

Answer:

step 4 I believe

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A circle is centered on point B, Points A, C and D lie on its circumference. if ABC measures 122 degrees, what does ADC measures

iren2701 [21]

Answer:

61°

Step-by-step explanation:

Angle extended to the circumference is half the angle at the centre

ADC = ½ ABC = ½ × 122 = 61

5 0

3 years ago

A hot air balloon is currently at a height of 900 feet. The balloon is descending by 10 feet minute. Find a function for the bal

mylen [45]

Answer:

$y=-10x+900$

Step-by-step explanation:

Let

x -----> the number of minutes

y ----> the balloon's height in feet

we know that

The linear equation in slope intercept form is equal to

$y=mx+b$

where

m is the slope

b is the y-coordinate of the y-intercept

In this problem we have that

The slope is equal to

$m=-10\ \frac{ft}{min}$ ----> is negative because is a decreasing function

$b=900\ ft$

Remember that the y-intercept is the value of y when the value of x is equal to zero

In this context the y-intercept is the height of the balloon when the time is equal to zero (initial height)

substitute

$y=-10x+900$

8 0

2 years ago

Use the quadratic formula to find the solutions to the equation x^2-3x+1=0

Gnom [1K]

Answer:

$x_{1}=\frac{+3+\sqrt{5} }{2}\\\\x_{2}=\frac{+3-\sqrt{5} }{2}$

Step-by-step explanation:

Using:

$x=\frac{-b+-\sqrt{b^{2}-4*a*c} }{2*a}$

we will have two solutions.

x^2-3x+1=0

So, a=1 b=-3 c=1

$x_{1}=\frac{+3+\sqrt{-3^{2}-4*1*1} }{2*1}\\\\x_{2}=\frac{+3-\sqrt{-3^{2}-4*1*1} }{2*1}$

We have two solutions:

$x_{1}=\frac{+3+\sqrt{5} }{2}\\\\x_{2}=\frac{+3-\sqrt{5} }{2}$

4 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

3 years ago