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3 years ago

If you were present 120 days out of 124 days what percent were you present

Mathematics

1 answer:

emmainna [20.7K]3 years ago

5 0

Answer:

~96.6

Step-by-step explanation:

120/124 is 30/31 or 60/62 or 96.6/99.82

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121 is 55% of what number

Wewaii [24]

Let x be the unknown number
x×55%=121
x=121÷55%
x=220
So 121 is 55% of 220

3 0

3 years ago

Mia's mother gives her $10 per week for allowance Mia gets an additional $3 for every hour she spends watching

nika2105 [10]

Answer:

10+3t=25

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Learning Task 3. Find the equation of the line. Do it in your notebook.

Wewaii [24]

Answer:

1) The equation of the line in slope-intercept form is $y = 5\cdot x +9$ . The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ . The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) The equation of the line in slope-intercept form is $y = 3\cdot x +4$ . The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ . The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ . The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

Step-by-step explanation:

1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: ( $m = 5$ , $x = -1$ , $y = 4$ )

$4 = (5)\cdot (-1)+b$

$4 = -5 +b$

$b = 9$

The equation of the line in slope-intercept form is $y = 5\cdot x +9$ .

Then, we obtain the standard form by algebraic handling:

$-5\cdot x + y = 9$

The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = 3$ , $y_{1} = 4$ , $x_{2} = -2$ , $y_{2} = 2$ )

$3\cdot m + b = 4$ (Eq. 1)

$-2\cdot m + b = 2$ (Eq. 2)

From (Eq. 1), we find that:

$b = 4-3\cdot m$

And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:

$-2\cdot m +4-3\cdot m = 2$

$-5\cdot m = -2$

$m = \frac{2}{5}$

And from (Eq. 1) we find that the y-Intercept is:

$b=4-3\cdot \left(\frac{2}{5} \right)$

$b = 4-\frac{6}{5}$

$b = \frac{14}{5}$

The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{2}{5}\cdot x +y = \frac{14}{5}$

$-2\cdot x +5\cdot y = 14$

The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: ( $m = 3$ , $b = 4$ )

$y = 3\cdot x +4$

The equation of the line in slope-intercept form is $y = 3\cdot x +4$ .

Then, we obtain the standard form by algebraic handling:

$-3\cdot x +y = 4$

The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -3$ , $y_{1} = 0$ , $x_{2} = 0$ , $y_{2} = 6$ )

$-3\cdot m + b = 0$ (Eq. 3)

$b = 6$ (Eq. 4)

By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:

$-3\cdot m+6 = 0$

$3\cdot m = 6$

$m = 2$

The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ .

Then, we obtain the standard form by algebraic handling:

$-2\cdot x +y = 6$

The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -1$ , $y_{1} = -2$ , $x_{2} = 5$ , $y_{2} = 3$ )

$-m+b = -2$ (Eq. 5)

$5\cdot m +b = 3$ (Eq. 6)

From (Eq. 5), we find that:

$b = -2+m$

And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:

$5\cdot m -2+m = 3$

$6\cdot m = 5$

$m = \frac{5}{6}$

And from (Eq. 5) we find that the y-Intercept is:

$b = -2+\frac{5}{6}$

$b = -\frac{7}{6}$

The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{5}{6}\cdot x +y =-\frac{7}{6}$

$-5\cdot x + 6\cdot y = -7$

The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

6 0

2 years ago

I need help on both of these please///:(

SCORPION-xisa [38]

14. The distance between the two points is 14.866.

Distance can be calculated with the following formula:

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(12-2)²+(5-(-6))²

d=√10²+11²

d=√100+121

d=√221

d=14.866

15. The distance between the two points is 20.248.

Use the same formula to find the distance.

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(4-(-3))²+(12-(-7))²

d=√7²+19²

d=√49+361

d=√410

d=20.248

4 0

3 years ago

Twenty-eight athletes participate in a 100-meter race. The time of each athlete is measured during the qualification round. The

andriy [413]

The average time of the 20 athletes who did not qualify for the final is 11.2 seconds .

<u>Step-by-step explanation:</u>

Here we have , Twenty-eight athletes participate in a 100-meter race. The time of each athlete is measured during the qualification round. The average time is 11 seconds. The 8 athletes with the fastest time qualified for the final. The average time of these 8 athletes is 10.5 seconds. Let's have equations for this:

$\frac{x_1+x_2+.....+x_2_8}{28} = 11$ where $x_1, x_2, ....., x_2_8$ are athletes

⇒ $x_1 + x_2 + .. + x_2_8 = 11(28)= 308$

⇒ $(x_1 + x_2 + .. + x_8 )+ (x_9+x_1_0+...+x_2_8) = 308$

⇒ $(\frac{x_1 + x_2 + .. + x_8}{8} )(8)+ (\frac{x_9+x_1_0+...+x_2_8}{20} )(20) = 308$

⇒ $10.5(8)+m(20) = 308$ , where m is average time of the 20 athletes who did not qualify for the final.

⇒ $84+m(20) = 308$

⇒ $20m= 224$

⇒ $m=11.2$

∴ The average time of the 20 athletes who did not qualify for the final is 11.2 seconds .

8 0

3 years ago