QUESTION: A survey of students in a homeroom class explored the
relationship between gender and participation in the school band. What
is a reasonable conclusion to draw from the data?*

[tex]x - \frac{3}{4} \geqslant - \frac{1}{8} [/tex]

[tex]x \geqslant - \frac{1}{8} + \frac{3}{4} [/tex]

Make both fractions to have same denominators by multiply 2 in 3/4 for both numerator and denominator.

[tex]x \geqslant - \frac{1}{8} + \frac{3 \times 2}{4 \times 2} \\ x \geqslant - \frac{ 1}{8} + \frac{6}{8} \\ x \geqslant \frac{5}{8} [/tex]

Hence, the solution is, "x is greater or equal to 5/8."

[tex] \large \boxed{x \geqslant \frac{5}{8} }[/tex]

Answer:

Add maybe

Step-by-step explanation:

add the 500+400+200 try that hope it helps have a great day

Answer:

The margin of error is 5. 00 

Answer:

Aug 16th

Step-by-step explanation:

Answer:

150 ft^2

Step-by-step explanation:

The formula is (30*10)/2 which is equal to 150.

Step-by-step explanation:

x+2x+6x=180

9x=180

x=20°

m<A = 20°

Answer:

m(arc CD) = 14.1 cm

Step-by-step explanation:

Measure of an arc of the circle = [tex]\frac{\theta}{360}(2\pi r)[/tex]

Here 'r' = radius of the circle

θ = Angle subtended by the arc at the center of the circle

From triangle OED,

sin(∠EOD) = [tex]\frac{ED}{OD}[/tex]

sin(∠EOD) = [tex]\frac{\frac{12.7}{2}}{9.06}[/tex]

                 = [tex]\frac{6.35}{9.06}[/tex]

                 = 0.7

m∠EOD = 44.498

              ≈ 44.5°

Angle subtended by the arc CD at the center = m∠COD

                                                                            = 2(44.5)

                                                                            = 89°

m(arc CD) = [tex]\frac{89}{360}(2\pi)(9.06)[/tex]

                 = 14.07

                 ≈ 14.1 cm

Answer:

[tex]y =5(3)^x[/tex]

Step-by-step explanation:

Given

The attached table

Required

Determine the equation of the table

The table represents an exponential function. So, we make use of

[tex]y = ab^x[/tex]

When x = 0, y= 5.

So:

[tex]y = ab^x[/tex] becomes

[tex]5 = ab^0[/tex]

[tex]5 = a*1[/tex]

[tex]5 = a[/tex]

[tex]a = 5[/tex]

When x = 1; y = 15

So;

[tex]y = ab^x[/tex] becomes

[tex]15 = a * b^1[/tex]

[tex]15 = a * b[/tex]

Substitute 5 for a

[tex]15 = 5 * b[/tex]

Divide both sides by 5

[tex]\frac{15}{5} = \frac{5 * b}{5}[/tex]

[tex]3 = b[/tex]

[tex]b = 3[/tex]

Substitute values for a and b in: [tex]y = ab^x[/tex]

[tex]y =5(3)^x[/tex]

Answer:

Y = x+3

Step-by-step explanation:

You want to start by figuring out how to get Y alone.

So to get it alone so you add +3 on both sides

Y - 3 = x

+3. +3

Positive 3 and Negative 3 cancel out making it zero.

Visual: 3 - 3 = 0

Since you cant take or add anything from x you leave it as + 3 so the answer should be.

Y = x + 3

I hope I could help :)

Answer:

Step-by-step explanation:

S, S, E.

Come on, do I have to spell it out for you?  :)

Answer:

(5, 5.25) , (10, 6.50) , (20, 9)

Step-by-step explanation:

As given ,

y = 0.25x + 4

Now,

if x = 5

⇒ y = 0.25(5) + 4

      = 1.25 + 4

     = 5.25

⇒ y = 5.25

∴ the ordered pair be (x, y) = (5, 5.25)

if x = 10

⇒ y = 0.25(10) + 4

      = 2.5 + 4

     = 6.50

⇒ y = 6.50

∴ the ordered pair be (x, y) = (10, 6.50)

if x = 20

⇒ y = 0.25(20) + 4

      = 5 + 4

     = 9

⇒ y = 9

∴ the ordered pair be (x, y) = (20, 9)

Given:

The polynomial function is

[tex]q(x)=6x^3+19x^2-15x-28[/tex]

Yuri thinks that [tex]\dfrac{3}{4}[/tex] is a root of the given function.

To find:

Why [tex]\dfrac{3}{4}[/tex] cannot be a root?

Solution:

We have,

[tex]q(x)=6x^3+19x^2-15x-28[/tex]

If [tex]\dfrac{3}{4}[/tex] is a root, then the value of the function at [tex]\dfrac{3}{4}[/tex] is 0.

Putting [tex]x=\dfrac{3}{4}[/tex] in the given function, we get

[tex]q(\dfrac{3}{4})=6(\dfrac{3}{4})^3+19(\dfrac{3}{4})^2-15(\dfrac{3}{4})-28[/tex]

[tex]q(\dfrac{3}{4})=6(\dfrac{27}{64})+19(\dfrac{9}{16})-\dfrac{45}{4}-28[/tex]

[tex]q(\dfrac{3}{4})=3(\dfrac{27}{32})+\dfrac{171}{16}-\dfrac{45}{4}-28[/tex]

[tex]q(\dfrac{3}{4})=\dfrac{81}{32}+\dfrac{171}{16}-\dfrac{45}{4}-28[/tex]

Taking LCM, we get

[tex]q(\dfrac{3}{4})=\dfrac{81+342-360-896}{32}[/tex]

[tex]q(\dfrac{3}{4})=\dfrac{-833}{32}\neq 0[/tex]

Since the value of the function at [tex]\dfrac{3}{4}[/tex] is not equal to 0, therefore, [tex]\dfrac{3}{4}[/tex] is not a root of the given function.

Answer:

Step-by-step explanation:

9 : 16 :: M : 32

Means of extremes = means of means

= 9 x 32 = 16 x M

= 288 = 16M

288/16 = M

18 = M

Answer:

oof

Step-by-step explanation:

Question- What method would you happen to be using to solve this problem?

Answer:B

Step-by-step explanation:m is the slope and m=-3

Answer:

this question is not complete. need another equation

Step-by-step explanation:

give another equation then I can solve

[tex] \frac{3}{8} x = \frac{17}{24} - \frac{1}{3} y \\ x = \frac{17}{9} - \frac{8}{9} y[/tex]

Answer:

273

Step-by-step explanation:

63 ÷ 30% = 210

210 + 63 = 273

therefore your total is 273

hope this is right, and that it helps :)

Answer:

Becky and her brother made 90+63=153 masks in total

Step-by-step explanation:

Percentages

Let's call x=number of masks Becky's brother made.

We know Becky made 30% less than the number of masks her brother made.

30% of x is calculated by:

30*x/100 = 0.3x

Now subtracting from x:

x - 0.3x = 0.7x

Becky made 63 carnival masks, this means that:

0.7x = 63

Dividing by 0.7:

x = 63/0.7 = 90

Becky's brother made 90 masks.

Becky and her brother made 90+63=153 masks in total

Answer:

810 is 100 times the size of 0.81

Answer:

[tex]t \ge h[/tex] and [tex]t + 0.5h \le 12.0[/tex]

Step-by-step explanation:

Given

[tex]t= thickness[/tex]

[tex]h =height[/tex]

From the first statement, we have that:

[tex]t\ mm \ge 0.1 * h\ cm[/tex]

Convert mm to cm

[tex]t\ * 0.1*cm \ge 0.1 * h\ cm[/tex]

[tex]t\ *0.1 \ge 0.1 * h[/tex]

Divide both sides by 0.1

[tex]t \ge h[/tex]

From the second statement, we have that:

[tex]Cost \le 12.2[/tex]

Substitute 0.2 + t + 0.5h for Cost

[tex]0.2 + t + 0.5h \le 12.2[/tex]

Collect Like Terms:

[tex]t + 0.5h \le 12.2-0.2[/tex]

[tex]t + 0.5h \le 12.0[/tex]

So, the inequalities are:

[tex]t \ge h[/tex] and [tex]t + 0.5h \le 12.0[/tex]

Answer: it's a, the answer

Step-by-step explanation:

Answer:

Sensitivity = 66.7%  (C)

specificity= 98.0%  (E)

positive predictive value = 80.0%  (F)

Negative predictive value = 96.0%  (D)

accuracy of the test  = 94.5%  (A)

Step-by-step explanation:

Given the data in the question;

                                    Disease Present              Disease Absent        Total

Test Positive                      24                                        6                        30

Test Negative                    12                                       288                     300

Total                                   36                                       294                    330

so A = 24, B = 6, C = 12 and D = 288

sensitivity = [A/(A+C)]×100 = [24/(24+12)]×100 = [24/36]×100

Sensitivity = 66.7%  (C)

specificity= [D/(D+B)]×100 = [288/(288+6)]×100 = [288/294]×100

specificity= 98.0%  (E)

positive predictive value = [A/(A+B)]×100 = [24/(24+6)]×100

= [24/30]×100

positive predictive value = 80.0%  (F)

Negative predictive value = [D/(D+C)]×100 = [288/(288+12)]×100

= [288/300]×100

Negative predictive value = 96.0%  (D)

accuracy of the test = [A+D/(A+B+C+D)]×100 = [24+288/(24+6+12+288)]×100

= [312/330]×100

accuracy of the test  = 94.5%  (A)

Nothing 33.3% (B)

Answer:

The answer is A, good luck on the test! :D

Step-by-step explanation:

Answer:

Step-by-step explanation:

A. In geometry, a cylinder is a three-dimensional shape with a circular base, ... So using the volume of cylinder formula we can find the volume of i

Answer: Y=2/3x+4

     ................................................

Answer:

im sooooo confused

Step-by-step explanation:

Compute the rightmost product:

[tex]\begin{bmatrix}1&2&0\\2&0&4\\0&4&-1\end{bmatrix}\begin{bmatrix}2\\2\\k\end{bmatrix}=\begin{bmatrix}1\times2+2\times2+0\times k\\2\times2+0\times2+4\times k\\0\times 2+4\times2+(-1)\times k\end{bmatrix}=\begin{bmatrix}6\\4k+4\\8-k\end{bmatrix}[/tex]

Then

[tex]\begin{bmatrix}2&2&k\end{bmatrix}\begin{bmatrix}6\\4k+4\\8-k\end{bmatrix}=2\times6+2\times(4k+4)+k\times(8-k)=-k^2+16k+20[/tex]

Use whatever methods you like to solve the remaining quadratic equation:

-k ² + 16k + 20 = 0   →   k = 8 ± 2√21