Enter your answer in the box.

Quadratic relations and comic sections unit test part 1

11. a. (0, +/- 6)

[tex] |\Omega|=2\cdot6=12\\
|A|=1\cdot2=2\\\\
P(A)=\dfrac{2}{12}=\dfrac{1}{6}\approx17% [/tex]

Answer: just add 8 + 8 and you will get 16

The projected sales total for the vendor at the concert is calculated by multiplying the average concession spending of $7.50 by the expected 10,000 fans, resulting in $75,000 (option c).

To calculate the projected sales total for the vendor at the concert, we need to multiply the average amount spent by each fan on concessions by the total number of fans expected to attend the event. Given that the average fan spends $7.50 on concessions and that there are 10,000 fans expected, the projected sales total can be found as follows:

Projected Sales Total = Average Spend per Fan x Total Number of Fans

Projected Sales Total = $7.50 x 10,000

Projected Sales Total = $75,000

Therefore, the correct answer is (c) $75,000.

The probability that a strawberry dessert contains whipped topping is approximately 69.2%

We are given that the probability a dessert contains strawberries (P(S)) is 26%, or 0.26, and the probability that a dessert contains both strawberries and whipped topping (P(S and W)) is 18%, or 0.18.

To find the probability that a randomly chosen strawberry dessert contains whipped topping (P(W|S)), we use the conditional probability formula:

[tex]P(W|S) = \frac{P(S and W)}{P(S)}[/tex]

Substituting the given values, we get:

[tex]P(W|S) = \frac{0.18}{0.26} \approx 0.6923[/tex]

Rounding to the nearest tenth of a percent we get the probability that a randomly chosen strawberry dessert contains whipped topping to be 69.2%.

Answer:

To find the zeros of a quadratic function, use the quadratic equation, [tex]x=\frac{-b \pm \sqrt{b^2-4ac} }{2a}[/tex]. We find that the eqautions with zeros at 1 and 4 are b) x² -5x + 4 and d) -2x² + 10x - 8.

Step-by-step explanation:

a) x² + x + 4 --

[tex]x = \frac{-1 \pm \sqrt{1^2-4*1*4} }{2*1}\\x=\frac{-1 \pm \sqrt{1-16}}{2}[/tex]

Because the discriminant (the value inside the square root) is negative, this equation does not have real zeros, so it is not the answer.

b) x² - 5x + 4 --

[tex]x = \frac{5 \pm \sqrt{(-5)^2-4*1*4}}{2*1} \\x=\frac{5 \pm \sqrt{25-16}}{2} \\x = \frac{5 \pm 3}{2}[/tex]

Now, we calculate the two zeros by adding and subtracting the 3.

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

The zeros of this function are 1 and 4, so it is included in our answer.

c) x² + 3x - 4 --

[tex]x = \frac{-3 \pm \sqrt{3^2-4*1*-4}}{2*1} \\x = \frac{-3 \pm \sqrt{9+16}}{2} \\x= \frac{-3 \pm 5}{2}\\\\x=\frac{-3+5}{2}=1\\x=\frac{-3-5}{2} = -4[/tex]

The zeros of this function are -4 and 1, so it is not the answer.

d) -2x² + 10x - 8 --

[tex]x = \frac{-10 \pm \sqrt{10^2-4*(-2)*(-8)} }{2*(-2)} \\x=\frac{-10 \pm \sqrt{100-64} }{-4} \\x = \frac{-10 \pm 6}{-4} \\\\x=\frac{-10 + 6}{-4} =1\\x = \frac{-10-6}{-4} =4[/tex]

The zeros of this function are 1 and 4, so it is included in our answer.

The graph of the function (f(x) = -14x - 2) is attached below and the two points from which line passes are (-0.143,0) and (0,-2).

Given :

Equation  --  f(x) = -14x - 2

The following steps can be used in order to sketch the graph of the given function:

Step 1 - Write the given function.

f(x) = -14x - 2

Step 2 - Now, evaluate the x-intercept of the above function.

0 = -14x - 2

x = -1/7

Step 3 - Now, determine the y-intercept of the given function.

f(x) = -2

Step 4 - Now, graph the equation of a line that passes through the points (-1/7,0) and (0,-2).

The graph of the function is attached below.

For more information, refer to the link given below:

https://brainly.com/question/14375099

Answer:

1 over 2 4 over 3 22 over 7 (12^3)

Step-by-step explanation:

Volume of sphere = (4/3)*pi*radius^3  

If the sphere has a diameter of 24 inches, then its radius is 12 inches.

The spherical fish bowl is half-filled; then, the volume of water inside the fish bowl is half of the volume of the bowl, that is,

volume of water = (1/2)*(4/3)*pi*radius^3  

Replacing with radius value and the given value for pi, we get:

volume of water = (1/2)*(4/3)*(22/7)*(12^3)

1. B. Counterclockwise

2. C. (30,401)

3. A. t=2(x-3)

4. C. She should have taken both the positive and negative square root

5. C. y=x^2+8x-25/8

6. D. Hyperbola

7. A. Graph A

Answer:

B,5/8+5r=8

Step-by-step explanation:

The exact value of the arc length of the curve is 149.4 units

How to determine the exact arc length of the curve

From the question, we have the following parameters that can be used in our computation:

[tex]y = \dfrac35x^\frac53 - \dfrac34x^\frac13 + 8[/tex]

Also, we have the interval to be

-1 ≤ x ≤ 27

This means that the x valus are

x = -1 to x = 27

The arc length of the curve can be calculated using

[tex]\text{Length} = \int\limits^a_b {\sqrt{1 + ((dy)/(dx))^2}} \, dx[/tex]

Recall that

[tex]y = \dfrac35x^\frac53 - \dfrac34x^\frac13 + 8[/tex]

So, we have

[tex]\dfrac{dy}{dy} = x^\frac{2}{3}-\dfrac{1}{4x^\frac{2}{3}}[/tex]

This means that

[tex]\text{Length} = \int\limits^{27}_{-1} {\sqrt{1 + (x^\frac{2}{3}-\dfrac{1}{4x^\frac{2}{3}})^2}} \, dx[/tex]

Using a graphing tool, we have the integrand to be

[tex]\text{Length} = \dfrac{12x^\frac{5}{3}+15\sqrt[3]{x}}{20}|\limits^{27}_{-1}[/tex]

Expand and evaluate

[tex]\text{Length} = 149.4[/tex]

Hence, the exact arc length of the curve is 149.4 units

0.8 or 80%.

The question is asking for the probability that a randomly chosen student is either a boy or a girl not wearing blue. There are 25 students in total, with 15 girls and 10 boys. Out of these, 5 girls and 4 boys are wearing blue. Therefore, the number of girls not wearing blue is 15 - 5 = 10 girls. Since all boys are considered in the probability, regardless of what they wear, we have 10 boys. So, we have 10 girls not wearing blue and 10 boys, totalling 20 students that match the criteria out of 25.

The probability can be calculated as follows:

( P(\text{{boy or girl not wearing blue}}) = \frac{{\text{{number of boys and girls not wearing blue}}}}{{\text{{total number of students}}}} = frac{{20}}{{25}} = 0.8 ) or 80%.

Therefore, the probability that a student picked at random is either a boy or a girl who is not wearing blue is 0.8 or 80%.

Answer:

c. I, II, and III

Step-by-step explanation:

The union of two sets includes all elements from one set and all elements from the second set.

For our sets, this means all elements of set A, which includes region I and region II.  It also means all elements of set B, which includes region III.

Thus the answer is regions I, II and III.

Answer:

C

Step-by-step explanation:

Just took the test

Answer:

The solutions are B. -4 and C. -5

Step-by-step explanation:

For a quadratic equation of the form [tex]ax^2+bx+c=0[/tex] the solutions are

[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

For [tex]\mathrm{}\quad a=3,\:b=27,\:c=60:\quad x_{1,\:2}=\frac{-27\pm \sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}[/tex]

[tex]x_1=\frac{-27+\sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}\\\\x_1=\frac{-27+\sqrt{9}}{2\cdot \:3}\\\\x_1=\frac{-27+3}{2\cdot \:3}\\\\x_1=\frac{-24}{6} = -4[/tex]

[tex]x_2=\frac{-27-\sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}\\\\x_2=\frac{-27-\sqrt{9}}{2\cdot \:3}\\\\x_2=\frac{-27-3}{2\cdot \:3}\\\\x_2=-\frac{30}{6} = -5[/tex]

│FV│ = │VD│

Being D the directrix. Given that the focus F is on the y-axis and the directrix is parallel to the x-axis, then the vertex V will also be on this axis, so h = 0.

As │FV│ = │VD│, then:

[tex]k = \frac{6-2}{2}[/tex], that is the middle point of the segment FD, so:

V(0,2)

Now │FV│= │p│= │2-(-2)│=4

Given that the vertex and focus are below the directrix, then the parabola open down, therefore: [tex]p\ \textless \ 0[/tex]

Lastly, the equation is:

[tex]x^{2} = -4(4)(y-2) = -16y+32[/tex]
[tex]y = -\frac{ x^{2} }{16} + 2[/tex]

The equation of a parabola with a focus at (0,-2) and a directrix of y=6 is x^2 = -16(y - 2).

To find the equation of a parabola with a focus at (0,-2) and a directrix of y=6, you need to use the standard form of the equation of a parabola that opens upwards or downwards. The general form of this type of parabola is  (x - h)^2 = 4p(y - k), where (h,k) is the vertex of the parabola, and p is the distance from the vertex to the focus (if the parabola opens upwards or downward) or to the directrix (if the parabola opens sideway).

Given that the focus is at (0,-2) and the directrix is at y=6, the vertex of the parabola will be located midway between them. The distance between the focus and directrix is 8 units, so the vertex will be 4 units from each, which puts the vertex at (0, 2). Therefore, h=0 and k=2.

Since the focus is below the directrix, our parabola opens downward, and the value of p is negative. The distance p is half the distance between the focus and directrix, so p=-4. Plugging these values into the general form, we get (x - 0)^2 = 4(-4)(y - 2), which simplifies to x^2 = -16(y - 2).

This is the equation that represents the desired parabola.

The children who collected more football than baseball cards are:

Ruso, Ryan , Monique , Jordan , Peter.

We are given a set of values as:

( Football cards,Baseball cards)       Letter           Name

     (10,60)                                             F                 Simian

     (10,90)                                             A                  Dan

     (20,40)                                             K                  Ken

     (30,40)                                             E                 Jimmy

     (30,60)                                             J                  Jason

     (40,40)                                             D                   Dyna

     (50,40)                                             C                   Ruso

     (60,20)                                             H                   Ryan

     (70,20)                                              I                  Monique

     (80,20)                                             G                  Jordan

     (80,50)                                              B                   Peter

Hence, children who collected more football then baseball cards are the one whose first value of the ordered pair is more than the other.

Hence, They are:

     (50,40)                                             C                 Ruso

     (60,20)                                             H                 Ryan

     (70,20)                                              I                 Monique

     (80,20)                                             G                 Jordan

     (80,50)                                              B                  Peter.