All categories

1 year ago

Five hundred sixth-grade students

Mathematics

1 answer:

viktelen [127]1 year ago

6 0

<h3>The ratio in three ways are 500 : 125 or 100 : 25 or 4 : 1</h3>

<em><u>Solution:</u></em>

Given that,

Five hundred sixth-grade students were surveyed

Which means,

sixth-grade students = 500

125 had traveled on an airplane

<em><u>What is the ratio of total sixth-grade students to the number of students who had traveled on an airplane?</u></em>

Therefore,

total sixth-grade students : students traveled on airplane = 500 : 125

<h3>Ratio = 500 : 125</h3>

Reduce to lowest terms,

Divide both sides by 5

<h3>Ratio = 100 : 25</h3>

Further reduce to simplest terms,

Divide both sides by 25

<h3>Ratio = 4 : 1</h3>

Thus the ratio in three ways are 500 : 125 or 100 : 25 or 4 : 1

You might be interested in

Which proportion satisfies the geometric mean (altitude) theorem for the triangle?

Degger [83]

D is the correct answer

3 0

2 years ago

Read 2 more answers

Mia’s work to find the slope of a trend line through the points (3, 10) and (35, 91) is shown below.

pochemuha

For this case the first thing you should know is that by definition the slope of the line is given by:
m = (y2-y1) / (x2-x1)
Where we have two ordered pairs:
P1 = (x1, y1)
P2 = (x2, y2)
We observe then that according to the formula, the error of mine was to invert the numerator with the denominator.
Answer:
C: Mia switched the numerator and the denominator.

5 0

1 year ago

Read 2 more answers

Suppose you borrowed $15,000 at a rate of 11.1% and must repay it in 5 equal installments at the end of each of the next 5 years

Elodia [21]

Answer:

The answer is $1,665.

Step-by-step explanation:

3 0

2 years ago

The general form of the equation of a circle is x2 + y2 + 42x + 38y − 47 = 0. The equation of this circle in standard form is (x

almond37 [142]

Step 1

we know that

The equation of a circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

$x^{2} +y^{2} +42x+38y-47=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+42x)+(y^{2}+38y)=47$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=47+21^{2}+19^{2}$

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=849$

Rewrite as perfect squares

$(x+21)^{2}+(y+19)^{2}=849$

The center of the circle is the point $(-21,-19)$

The radius of the circle is $\sqrt{849}\ units$

<u>The answer Part a) is</u>

The equation of the circle in standard form is equal to

$(x+21)^{2}+(y+19)^{2}=849$

<u>The answer Part b) is</u>

The center of the circle is the point $(-21,-19)$

<u>The answer Part c) is</u>

The radius of the circle is $\sqrt{849}\ units$

Let's verify each case to determine the solution of the second part of the problem

Step 2

we have

$x^{2} +y^{2} +60x+14y+98=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+60x)+(y^{2}+14y)=-98$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=-98+30^{2}+7^{2}$

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=851$

Rewrite as perfect squares

$(x+30)^{2}+(y+7)^{2}=851$

The radius of the circle is $\sqrt{851}\ units$

therefore

This circle does not have the same radius of the circle above

Step 3

we have

$x^{2} +y^{2} +44x-44y+117=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+44x)+(y^{2}-44y)=-117$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=-117+22^{2}+22^{2}$

$(x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=851$

Rewrite as perfect squares

$(x+22)^{2}+(y-22)^{2}=851$

The radius of the circle is $\sqrt{851}\ units$

therefore

This circle does not have the same radius of the circle above

Step 4

we have

$x^{2} +y^{2} -38x+42y+74=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-38x)+(y^{2}+42y)=-74$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=-74+19^{2}+21^{2}$

$(x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=728$

Rewrite as perfect squares

$(x-19)^{2}+(y+21)^{2}=728$

The radius of the circle is $\sqrt{728}\ units$

therefore

This circle does not have the same radius of the circle above

Step 5

we have

$x^{2} +y^{2} -50x-30y+1=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-50x)+(y^{2}-30y)=-1$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-50x+25^{2})+(y^{2}-30y+15^{2})=-1+25^{2}+15^{2}$

$(x^{2}-50x+25^{2})+(y^{2}-30y+15^{2})=849$

Rewrite as perfect squares

$(x-25)^{2}+(y-15)^{2}=849$

The radius of the circle is $\sqrt{849}\ units$

therefore

This circle has the same radius of the circle above

therefore

<u>The answer is</u>

$x^{2} +y^{2} -50x-30y+1=0$ -----> has the same radio that the circle above

7 0

1 year ago

Alvin’s first step in solving the given system of equations is to multiply the first equation by 2 and the second equation by –3

skelet666 [1.2K]

The answer is C, one solution 0x+15.5y=144. Hope this helps!!

8 0

1 year ago

Read 2 more answers