Will mark brainliest and 25 points please help c:

ABBBAAA what is the a to b ratio?

Ede4ka [16]

there are 4 A and 3 B.

so the ratio is 4 : 3

3 0

3 years ago

I'm stuck on algebra.

Rainbow [258]

Point slope form : y-y1 = m(x-x1)
slope intercept form : m=(y-y1)/(x-x1)
standard form : Ax + By = C

just pick any random points on a graph and plug the same points into all three forms.

7 0

3 years ago

Read 2 more answers

There are 92,746 seats available for fans at UGA's Sanford Stadium. Auburn University's Jordan-Hare Stadium has seating for 87,4

Anvisha [2.4K]

Given :

Number of seats at UGA's Sanford Stadium, $n_1 = 92,746$ .

Number of seats at Auburn University's Jordan-Hare Stadium, $n_2=87,451$ .

To Find :

How many more seats are there at UGA's stadium.

Solution :

Extra seats at UGA's stadium = seats at UGA's Sanford Stadium - seats at

Auburn University's Jordan-Hare Stadium

= 92,746 - 87,451

= 5295

Hence, this is the required solution.

4 0

2 years ago

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

In which

x is the number of sucesses

$
e = 2.71828$ is the Euler number

$\mu$ is the mean in the given interval.

Binomial 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.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

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

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

8 0

3 years ago

A field is in the shape of a rectangle 5/6 mile long and 3/4 mile wide. What is the area of the field? *

Lynna [10]

Area = length x width

Area = 5/6 x 3/4

= (5x3) / (6x4)

= 15/24

= 5/8 square miles

7 0

3 years ago