1 year ago

Match with the sum of the monomial and binomial

Mathematics

1 answer:

Margarita [4]1 year ago

5 0

Answer:

The answer to your question is below

Step-by-step explanation:

Just cancel the parentheses and simplify

Result Addition

5xy² + 2x²y ⇒ (5xy² - x²y) + 3x²y = 5xy² + 2x²y

-3x²y - 2xy² ⇒ x²y + (-2xy² - 4x²y) = -3x²y - 2xy²

3x²y - 2xy² ⇒ 2x²y + (x²y - 2xy²) = 3x²y - 2xy²

5x²y + xy² ⇒ 4x²y + (x²y + xy²) = 5x²y + xy²

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A game is played using one die. If the die is rolled and shows 1, the player wins $5. If the die shows any number other than 1,

mash [69]

Answer:

1/5

Step-by-step explanation:

i had a similar question

For each roll you start with paying 2 dollars and you only with 10 dollars one out of 6 rolls (on average).

So the cost for one play is 2 dollars and your win is 10/6.

Value is -2+10/6=-1/3 dollars

So you lose 1/3 dollars on average with each game

since you have no limited rolls u put 1/5

this from another question but both same just different numbers

6 0

2 years ago

A set of data has a mean of 56.1. The data follows a normal distribution curve and has a standard deviation of 8.2. Find the pro

Whitepunk [10]

Given that mean=56.1 and standard deviation=8.2, P(x>67.5) will be found as follows:
The z-score is given by:
z=(x-μ)/σ
thus the z-score will be given by:
z=(67.5-56.1)/8.2
z=11.4/8.2
z=1.39
thus
P(z=1.39)=0.9177
thus:
P(x>67.5)=1-P(z>0.9177)
=1-0.9177
=0.0823
Answer: A. 0.0823

7 0

1 year ago

Read 2 more answers

The typical college freshman spends an average of μ=150 minutes per day, with a standard deviation of σ=50 minutes, on social me

Rashid [163]

Answer: The proportion of students spending at least 2 hours on social media equals 0.7257 .

Step-by-step explanation:

Given : The typical college freshman spends an average of μ=150 minutes per day, with a standard deviation of σ=50 minutes, on social media.

The distribution of time on social media is known to be Normal.

Let x be the number of minutes spent on social media.

Then, the probability that students spending at least 2 hours (2 hours = 120 minutes as 1 hour = 60 minutes) on social media would be:

$P(x\geq120)=1-P(x$