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2 years ago

Which are equivalent to −6(b+2)+8 chose all that apply −6b+2+8 −6b−4 None of the above

Mathematics

1 answer:

xeze [42]2 years ago

4 0

None of the above are equivalent to -6(b+2)+8

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At a football game, your team lost 55 yards on the first play and then 77 yards on the second play. Which expression gives the t

Aleks04 [339]

The answer is the third one (C?) because if you started at 0, you are going below it since you are losing yards, causing you to have less and less yards. Adding two negatives gives you a lesser amount.

4 0

3 years ago

Read 2 more answers

Will, Mike, and Sue went to dinner. Will paid 1/3 of the dinner bill. Micah and Sue paid in the ratio 2:3. If Sue paid $6 more t

raketka [301]

Let's call their parts w,m and s

if will paid 1/3, then m+s=2w (they paid 2/3, which is twice as much as will did)

Now, we know that:
Micah and Sue paid in the ratio 2:3.
this means the 3m=2s

and m=2/3s

Again:
m+s=2w

and we substitute m:
2/3s+s=2w
5/3s=2w// multiply both sides by 3
5s=6w

we also know that s=w+6 (from the last sentence) so we substitute:

5(w+6)=6w
5w+30=6w
30=w

so, Will paid 30, Sue paid 36 (six more than him), Mike paid 24 (24:36 is the same ratio as 2:3, you can check this by dividing both 24 and 36 by 12: you have 2 and 3)

and the total was 30+36+24=90.

3 0

3 years ago

Brandon has a playlist with 83 hip-hop and pop songs. There are 15 more hip-hop songs than pop songs. How many pop songs are on

kogti [31]

Using a system of equations, it is found that there are 34 pop songs are on the playlist.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

Variable x: Number of hip-hop songs.

Variable y: Number of pop songs.

Brandon has a playlist with 83 hip-hop and pop songs, hence:

x + y = 83.

There are 15 more hip-hop songs than pop songs, hence:

x = 15 + y.

Replacing the second onto the first equation:

x + y = 83.

15 + y + y = 83.

2y = 68.

y = 34.

There are 34 pop songs are on the playlist.

More can be learned about a system of equations at brainly.com/question/24342899

6 0

2 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago

Give a real life example of two variables that very directly

Snowcat [4.5K]

Answer:

the amount of products a company sells directly affects the amount of money it has

7 0

3 years ago