which of the expressions doe not contain any like terms? a. 2x2+3x-4x+2 b. xy+yz-17 c. x2+x2-2x-x2 d. 4+8+y

Answers

Answer 1

Answer: b. does not contain any like terms, because xy, yz, and -17, do not share any common variables.

Hope this helps!

Answer 2

Answer: The answer is B it does not have any of the other terms

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Use the given data to find the minimum sample size required to estimate a population proportion or percentage. margin of error: 0.040.04; confidence level 9595%; modifyingabove p with caretp and modifyingabove q with caretq unknown

Answers

The minimum sample size required to estimate a population proportion or percentage is 306.

What is random sampling?

In statistics, a simple random sample is a subset of individuals chosen from a larger set in which a subset of individuals are chosen randomly, all with the same probability. It is a process of selecting a sample in a random way.

In order to determine the minimum sample size required to estimate a population proportion or percentage, we will use the following formula:

n = (z×p×q)/m²

where is the minimum sample size, z is the z-score corresponding to the desired confidence level, p is the population proportion, q is 1-p, and m is the desired margin of error.

In this case, the confidence level is 95%, so the corresponding z-score is 1.96. Since we don't know the population proportion, we will use the symbol p and q to represent it. Therefore, the formula becomes:

n = (1.96×p×q)/(0.04)²

To determine the minimum sample size, we need to determine the value of p and q. Since p + q = 1, if we set p to 0.5, then q will also be 0.5. Therefore, the minimum sample size is:

n = (1.96×0.5×0.5)/(0.04)² = 306.25

≈ 306

Therefore, the minimum sample size required to estimate a population proportion or percentage is 306.

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Minimum required sample size for a desired margin of error and confidence level when it is a proportion problem: n = (z2÷margin of error2)*p-hat*q-hat

The maximum value of p-hat*q-hat occurs where p-hat = .5 (found by taking the derivative of (p-hat)*(1-p-hat) and setting it equal to 0 to find the maximum. n = ( 2.5762( for 99% confidence interval)÷.0482 )*.5*.5 = 720.028 or 721

Express 0.77230.77230, point, 7723 as a fraction

Answers

Answer:

7723/10000

Step-by-step explanation:

find the minimum value of p=10x+26y the constraints are x+y less than or equal to 6, 5x+y greater than or equal to 10, x+5y greater than or equal to 14

Answers

Answer:

Minimum value of $p=10x+26y$ is 80 at (1.5,2.5)

Step-by-step explanation:

We are given

The objective function is, Minimize $p=10x+26y$

With the constraints as,

$x+y\leq 6\n5x+y\geq 10\nx+5y\geq 14$

So, upon plotting the constraints, we see that,

The boundary points of the solution region are,

(1,5), (1.5,2.5) and (4,2).

So, the minimum values at these points are,

Points $p=10x+26y$

(1,5) $p=10x* 1+26* 5$ i.e. p = 140

(1.5,2.5) $p=10* 1.5+26* 2.5$ i.e. p= 80

(4,2) $p=10* 4+26* 2$ i.e. p = 92

Thus, the minimum value of $p=10x+26y$ is 80 at (1.5,2.5).

Consider the 1-to10^19 scale on which the disk of the Milky Way Galaxy fits on a football field. On this scale, how far is it from the sun to the alpa centauri (real distance:4.4 light-years) How big is the sun itself on this scale? Compare the sun's size on this scale to the actual size of a typical atom (about 10^-10 m in diameter).

Answers

Answer:

Sun is $0.4162708$ cm away from alpha century.

Sun is $1.391016*10^(-10)$ m.

Sun is 1.391016 time the size of an atom at this scale.

Step-by-step explanation:

Light year is a measure of distance. It is the distance light travels in an year.

Light year = $9.4607*10^(12)$ km

So 4.4 light years = $4.4*9.4607*10^(12)$ km

$41.62708*10^(12)$ km

Lets scale this down to the level of $1*10^(-19)$

$41.62708*10^(12)*1*10^(-19)$ km

= $41.62708*10^(-7)$ km

Change the units to centimeters:

$41.62708*10^(-7)*1*10^(5)$ cm

= $41.62708*10^(-2)$ cm

= $0.4162708$ cm

Therefore on the new scale sun is $0.4162708$ cm away from alpha century.

Diameter of the sun is 1.391016 million km

Lets change Sun's diameter to the new scale:

$1.391016*10^(6) *10^(-19)$ km

= $1.391016*10^(-13)$ km

Lets change kilometers in to meters:

$1.391016*10^(-13)*10^(3)$ m

= $1.391016*10^(-10)$ m

Therefore, sun is $1.391016*10^(-10)$ m

and an atom is $1*10^(-10)$

Therefore the sun is 1.391016 time the size of an atom at this scale.

Final answer:

On a 1-to-10^19 scale, the distance from the Sun to the Alpha Centauri is about 44 cm. On this same scale, the Sun itself would have a diameter of about 150 picometers, which is larger than a typical atom.

Explanation:

The 1-to-10^19 scale means for every actual meter in space, we represent 10^19 meters on our model. The Alpha Centauri is 4.4 light-years away from the sun. Considering 1 light-year equals to approximately 9.46x10^15 meters, the real distance from the sun to Alpha Centauri is about 4.16x10^16 meters. So, on the scale, this is about 0.44 meters or 44 cm.

The Sun's real size, with a diameter of 1.5 million kilometers or 1.5x10^9 meters, is represented as 1.5x10^-10 meters or 150 picometers on the scale. This is much bigger than an actual atom, which has a diameter of 0.1 to 0.5 nanometers or 100 to 500 picometers. Hence, on this scale, the Sun would be larger than a typical atom.

Learn more about Scale Representation here:

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Sasha finished mowing lawns at 4:15 PM. It took 1 3/4h to mow the first lawn. The second and third lawns she mowed each took 1 1/5 h. She took one 45-min break. When did Sasha begin mowing the first lawn? A. 10.05 A.M. B. 11:05 A.M. C. 11:21 A.M. D. 12:21 P.M.

Answers

The time it took to mow all the lawns including the break he took was 1 3/4+1 1/5+1 1/5+3/4=
2 1/2+2 2/5= 4 9/10 which is 4 hours and 54 minutes and if he got done at 4:15pm then he started 4 hours and 54 minutes before at
C. 11:21 A.M.

Which of these is an example of a literal equation?A.3x – 4y
B.12 = 9 + 3x
C.6 + 30 = 62
D.ax – by = k

Answers

ANSWER

$ax-by=k$ is an example of literal equation.

EXPLANATION

A literal equation is an equation in which letters or variables are used to represent real values.

A literal equation consists of at least two letters or variables.

The first option consists of two variables but it is not an equation. It is just an expression.

The second option is not a literal equation because it consists of only one variable. This is just a linear equation in one variable. But a literal equation should have at least two variables or letters.

As for the third option, it does not even contain a variable or letter.

Choice D. represents a literal equation

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