MATHEMATICS MIDDLE SCHOOL

In your class, 2 out of every 20 students are left-handed. What percent of students in your class are left handed?

Answers

Answer 1

Answer: $(left-handet)/(all students) = (2)/(20) = (10)/(100) =$ 10%

Answer : 10%

Answer 2

Answer:

Hey, Tyty1235!

2/20=left handed

1/10=left handed

1*10

10*10

10/100=10%

I really hope this helps;)

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X+y=20 is it linear? is it proportional?

Answers

Linear not proportional good luck

Nope it's not because if it was proportional it would be like y=kx. K stands for a number or the unit rate or the constant of proportionality!

When dividing 878 by 31, a student finds a quotient of 28 with a remainder of 11. Check the students work, and use the check to find the error in the solution

Answers

There is an error in the student's work, the remainder should be 10

Further explanation

The division operation is one of the basic arithmetic operations

Common signs used are $\large{\boxed{:}}}$ or $\large{\boxed{\slash}}$

The division is written by placing the numerator above the denominator with a horizontal line between them or use a slash with the numerator's position and the denominator is parallel

The division is actually an iterative subtraction operation (as many as the number of divisors)

Division is the opposite of multiplying

The general form of division can be stated by:

$\large{\boxed{\bold{divident:divisor=quotient+remainder}}$

example:

7: 3 = 2 + 1

7 = dividend
3 = divisor
2 = quotient
1 = remainder

Division operation statement:

"7 divided by 3 equals 2 remainder 1"

The form of division can also be expressed in terms of fractions

In the example above, the form of mixed fractions :

$\displaystyle (7)/(3) =2(1)/(3)$

Mixed fractions consist of whole number and proper fractions

From the task

(see also the attached image)

When dividing 878 by 31, a student finds a quotient of 28 with a remainder of 11.

Division operation statement:

"878 divided by 31 equals 28 remainder 11

Arithmetic operations :

878 : 31 = 28 + 11

878 = dividend
31 = divisor
28 = quotient
11 = remainder

Because Division is the opposite of multiplying, so we check the solution with multiplying

divident - (divisor x quotient) = remainder

878 - (31 x 28) = 11

878 - 868 = 11

10 ≠ 11

So students make mistakes in subtraction operations,remainder should be 10

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Keywords: division, mixed fraction, dividend , divisor , quotient, remainder

878/3 is 28 with a remainder of 10! The error is when they subtracted 258-248 wrong

Tyler simplified the expression. His procedure is shown below.

Answers

Answer:

B:Both powers should be in the denominator with positive exponents.

Step-by-step explanation:

We are given that Tyler simplified an expression

$x^(-3)y^(-9)$

We have to find an error which Tyler makes in her simplification

Step 1: $x^(-3)y^(-9)$

Step 2: $(1)/(x^3)\cdot y^(-9)$

Step 3: $(1)/(x^3y^9)$

Hence, Tyler's error is both exponents should be denominator with positive exponents.

Answer:B:Both powers should be in the denominator with positive exponents.

Answer:

Step-by-step explanation:

Tyler correctly rewrote x^(-3) as ------- , but

x³

he incorrectly wrote y^(-9) as ---------

y^(-9)

Both x^9 and y^9 should be in the denominator. This corresponds to the 2nd given possible answer.

How do you find the volumke of a cylinder

Answers

V=πr2h is how you find it

Question : What is the constant of proportionality in the equation? ----------------------------------------------------------------------------------------------------------------------------Choices: y = 5 x A) 0 B) 1 C) 4 D) 5

Answers

Answer:

1 1/2

Step-by-step explanation:

y = 1 1/2x

The standard equation of direct variation is

y = kx,

where k = constant of proportionality.

Here you have

y = 1 1/2x,

so 1 1/2 is k, the constant of proportionality.

Answer: 1 1/2

Hopes this help please do me as brainlist and hit that thanks.

The equation 7^2=a^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A, in astronomical units, AU. If planet Y is k times the mean distance from the sun as planet X, by what factor is the orbital period increased?

Answers

Answer:

The period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

Step-by-step explanation:

The equation $T ^ 2 = a ^ 3$ shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?

For the planet Y:

$T_y ^ 2 = a_y ^ 3$

For planet X:

$T_x ^ 2 = a_x ^ 3$

To know the factor of aumeto we compared $T_x$ with $T_y$

We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:

$a_y ^ 3 = (a_xk) ^ 3$

$(T_y ^ 2)/(T_x ^ 2)=(a_y ^ 3)/(a_x^ 3)\n\n(T_y ^ 2)/(T_x ^ 2)=((a_xk)^3)/(a_x ^ 3)\n\n(T_y^ 2)/(T_x^ 2)=\frac{k ^ {3}a_(x)^ 3}{a_(x)^ 3}\n\n(T_(y)^ 2)/(T_(x)^ 2)=k ^ 3\n\nT_(y)^ 2 = T_(x)^(2)k^(3)\n\nT_(y) =k^{(3)/(2)}T_x$

Finally the period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

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