$5.52 for 24 soda cans

A, B, and C are polynomials, where A = n, B = 2n + 6, and C = n2 – 1. What is AB – C in simplest form? A=–n2 + 3n + 5 B=n2 + 6n

slava [35]

Answer: B: n^2+6n+1

Step-by-step explanation:

A=n

B=2n+6

C=n^2-1

AB-C

n(2n+6)-n^2-1

2n^2+6n-n^2+1

n^2+6n+1

4 0

2 years ago

Read 2 more answers

The weights of cars passing over a bridge have a mean of 3,550 pounds and standard deviation of 870 pounds. Assume that the weig

Annette [7]

Answer:

0.24315

Step-by-step explanation:

Using the z score formula to solve this question

z = (x - μ) / σ,

Such that:

x = raw score

μ = population mean

σ = population standard deviation.

From the question:

x = 3000

μ = 3550

σ = 870

z = (3000 - 3550) / 870

z = -550/870

z = -0.6962

Using the z score table as well as probability calculator(as requested in the question to find the z score)

The probability of having less than 3000 is obtained as:

P(x<3000) = 0.24315

7 0

2 years ago

When a decision maker is faced with several alternatives and an uncertain set of future events, s/he uses __________ to develop

Step2247 [10]

Answer:

Option b: a decision analysis

Step-by-step explanation:

Option b: a decision analysis

Decision analysis is referred to that the strategic marketing planning process that consists of a systematic visual approach. it facilitates the use of all tools that can be needed to have a discussion on the decision process.

An approach like a diagrammatic representation of available resources, choices are mentioned on influence to easily draw the decision.

4 0

2 years ago

Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can

jek_recluse [69]

Answer:

0.278

Step-by-step explanation:

Given that Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can be used. B box has three keys but only one can be used. C box has two keys but none of them can be used.

Each box is equally likely to be selected.

In other words

$P(A) = P(B) = P(C)=\frac{1}{3}$

If A is selected then probability of winning is using the correct key out of two keys i.e. 0.5

If B is selected then probability of winning is using the correct key out of three keys i.e. 0.333

If c is selected then probability of winning is using the correct key out of two keys i.e. 0.00

So the probability that Nuri can win the car

= $\frac{1}{3} *0.5+\frac{1}{3} *0.333+\frac{1}{3} *0\\= 0.278$

6 0

2 years ago

Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ou

monitta

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

B) 0.6796

C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = $P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )$

= P ( Z > 2.33 ) = 0.0099

= 0.9%

B) percentage of oranges weighing less than 8.7 ounces

P( x < 8.7 ) = $P ( \frac{x - u}{ std} > \frac{8.7-8}{1.5} )$

= P ( Z < 0.4667 ) = 0.6796

= 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = $P (\frac{6.2-8}{1.5} < \frac{x - u}{ std} < \frac{7-8}{1.5} )$

= P ( -1.2 < Z < -0.66 )

= Ф ( -0.66 ) - Ф(-1.2) = 0.13956

7 0

2 years ago