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3 years ago

13

It must be least to greatest

1 answer:

stiv31 [10]3 years ago

4 0

-5 1/3, 2 1/2, 3.75 that is all i can see in the picture

You might be interested in

How many solutions does the system of equations below have?

soldier1979 [14.2K]

Answer:

One solution

Step-by-step explanation:

5x + y = 8

15x + 15y = 14

Lets solve using substitution, first we need to turn "5x + = 8" into "y = mx + b" or slope - intercept form

So we solve for "y" in the equation "5x + y = 8"

5x + y = 8

Step 1: Subtract 5x from both sides.

5x + y − 5x = 8 − 5x

Step 2: 5x subtracted by 5x cancel out and "8 - 5x" are flipped

y = −5x + 8

Now we can solve using substitution:

We substitute "-5x + 8" into the equation "15x + 15y = 14" for y

So it would look like this:

15x + 15(-5x + 8) = 14

Now we just solve for x

15x + (15)(−5x) + (15)(8) = 14(Distribute)

15x − 75x + 120 = 14

(15x − 75x) + (120) = 14(Combine Like Terms)

−60x + 120 = 14

Step 2: Subtract 120 from both sides.

−60x + 120 − 120 = 14 − 120

−60x = −106

Divide both sides by -60

$\dfrac{ -60x }{ -60 } = \dfrac{ -106 }{ -60 }$

Simplify

$x = \dfrac{ 53 }{ 30 }$

Now that we know the value of x, we can solve for y in any of the equations, but let's use the equation "y = −5x + 8"

$\mathrm{So\:it\:would\:look\:like\:this:\ y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}$

$\mathrm{Now\:lets\:solve\:for\:"y"\:then}$

$y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}$

$\mathrm{Express\: -5 \times \dfrac{ 53 }{ 30 }\:as\:a\:single\:fraction}$

$y = \dfrac{ -5 \times 53 }{ 30 } +8$

$\mathrm{Multiply\:-5 \:and\:53\:to\:get\:-265 }$

$y = \dfrac{ -265 }{ 30 } +8$

$\mathrm{Simplify\: \dfrac{ -265 }{ 30 } \:,by\:dividing\:both\:-265\:and\:30\:by\:5} }$

$y = \dfrac{ -265 \div 5 }{ 30 \div 5 } +8$

$\mathrm{Simplify}$

$y = - \dfrac{ 53 }{ 6 } +8$

$\mathrm{Turn\:8\:into\:a\:fraction\:that\:has\:the\:same\:denominator\:as\: - \dfrac{ 53 }{ 6 }}$

$\mathrm{Multiples\:of\:1: \:1,2,3,4,5,6}$

$\mathrm{Multiples\:of\:6: \:6,12,18,24,30,36,42,48}$

$\mathrm{Convert\:8\:to\:fraction\:\dfrac{ 48 }{ 6 }}$

$y = - \dfrac{ 53 }{ 6 } + \dfrac{ 48 }{ 6 }$

$\mathrm{Since\: - \dfrac{ 53 }{ 6 }\:have\:the\:same\:denominator\:,\:add\:them\:by\:adding\:their\:numerators}$

$y = \dfrac{ -53+48 }{ 6 }$

$\mathrm{Add\: -53 \: and\: 48\: to\: get\: -5}$

$y = - \dfrac{ 5 }{ 6 }$

$\mathrm{The\:solution\:is\:the\:ordered\:pair\:(\dfrac{ 53 }{ 30 }, - \dfrac{ 5 }{ 6 })}$

So there is only one solution to the equation.

5 0

1 year ago

This year, Mrs. Fisher has 9 students who are new to the school. The number of new students represents 7.5% of all her students.

Lelu [443]

$7.5\% \text { of } x = 9 \text { students}$

$0.75 \times x = 9 \text { students}$

$0.75x = 9$

5 0

2 years ago

at a certain conference of 100 people there are 29 indian women and 23 indian men. out of these indian people, 4 are doctors and

tangare [24]

Answer:

4 women doctors

Step-by-step explanation:

because if 24 are men or doctor the 4 gotta be women doctor

3 0

2 years ago

A repeated-measures study comparing two treatments with n = 4 participants produces md = 2 and ss = 75 for the difference scores

valentinak56 [21]

The estimated standard error for the sample mean difference is 2.5 .

According to the question

A repeated-measures study comparing two treatments

n = 4

MD(mean difference) = 2

SS (sum of square) = 75

Now,

error for the sample

Formula for standard error

$S^{2} = \frac{SS}{n-1}$

by substituting the value

$S^{2} = \frac{75}{4-1}$

$S^{2} = \frac{75}{3}$

S² = 25

S = 5 (s is never negative)

Standard error of the estimate for the sample mean difference

As

The standard error of the estimate is the estimation of the accuracy of any predictions.

The formula for standard error of the mean difference

standard error of the mean difference = $\frac{standard\\\ error}{\sqrt{n} }$

standard error of the mean difference = $\frac{5}{\sqrt{4} }$

standard error of the mean difference = 2.5

Hence, the estimated standard error for the sample mean difference is 2.5 .

To know more about estimated standard error here:

brainly.com/question/14524236

#SPJ4

5 0

1 year ago

Un jardín rectangular de 50 cm de largo por 34 m de ancho está rodeado por un camino de arena uniforme.Halla la anchura de dicho

Anastasy [175]

Answer:

5.85 m

Step-by-step explanation:

The width of the sand road can be calculated knowing its area and the dimensions of the rectangular garden as follows:

$A_{g} = a.b$

<u>Where:</u>

Ag: is the area of the rectangular garden

a: is the length of the rectangular garden = 50 cm = 0.5 m

b: is the width of the rectangular garden = 34 m

$A_{s} = 540 m^{2}$

<u>Where</u>:

As: is the area of the sand road

The relation between the area of the sand road and the area of the rectangular garden is the following:

$A_{s} + A_{g} = (a+2x)*(b+2x)$

$540 m^{2} + 0.5m*34m = ab + 2ax + 2bx + 4x^{2}$

$557 m^{2} = ab + 2x(a + b) + 4x^{2}$

$557 m^{2} - 17m^{2} - 2x(34.5 m) - 4x^{2} = 0$

$540 - 69x - 4x^{2} = 0$

By solving the above equation for x we have two solutions:

x₁ = -23.10 m

x₂ = 5.85 m

Taking the positive value, we have that the width of the sand road is 5.85 m.

I hope it helps you!

7 0

2 years ago