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riadik2000 [5.3K]
3 years ago
15

The spinner shown has congruent sections. The spinner is spun 63 times. What is a reasonable prediction for the number of times

the spinner will land on a section that is shaded red?​
Mathematics
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

36 times

Step-by-step explanation:

Given

Red = 4

Total = 7

n = 63

See attachment

Required

Predicted number of times to land on red

First, we calculate the probability of landing on red

Pr = \frac{Red}{Total}

Pr = \frac{4}{7}

The predicted number of red is the =n calculated as:

Red = Pr * n

Red = \frac{4}{7}*63

Red = 4 * 9

Red = 36

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Answer:

<h2>$26.25</h2>

<em><u>Solving steps:</u></em>

<em>Question:</em> <u>Sam had some money in his pocket, and he found another $6. 50 in his dresser drawer. He then had a total of $19. 75. Let p represent the amount of money Sam had in his pocket. Which equation can you use to find the amount of money Sam had in his pocket? How much money did Sam have in his pocket?.</u>

<em>Find: </em><em> </em><u>How much money did Sam have in his pocket?.</u>

<em>Solution:</em><em> </em>Let the equation be

<h3><em>=> P = T </em><em>+</em><em>F</em></h3>

<u>p represent amount of money</u>

<u>p represent amount of moneyt represent total</u>

<u>p represent amount of moneyt represent totalf represent money found</u>

<h3><em>=> P = T </em><em>+</em><em> </em><em>F</em></h3>

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<h3><em>=> P = $19.75 </em><em>+</em><em> </em><em>$6.50</em></h3>

add<u> 19.75 from 6.50 </u>

<h3><em>=> P = </em><em> </em><em>26.25</em></h3>

<em><u>THEREFORE THE AMOUNT OF MONEY </u></em><em><u>SAM</u></em><em><u> HAVE IN HIS POCKET</u></em><em><u> IS ABOUT</u></em><em><u> </u></em><em><u> </u></em><em><u>$</u></em><em><u>26.25</u></em>

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1. The height of a triangle is 6 m more than its base. The area of the triangle is 56 m². What is the length of the base? Enter
Elodia [21]
Answers:
1. 8 m 
2. 17 m
3. 7 cm
4. 2 s

Explanations:

1. Let x = length of the base
          x + 6 = height of the base

    Then, the area of the triangle is given by

    (Area) = (1/2)(base)(height)
       56 = (1/2)(x)(x + 6)
       56 = (1/2)(x²  + 6x) 
     
    Using the symmetric property of equations, we can interchange both sides      of equations so that 

    (1/2)(x²  + 6x) = 56
    
    Multiplying both sides by 2, we have
   
    x² + 6x = 112
    
    The right side should be 0. So, by subtracting both sides by 112, we have 

    x² + 6x - 112 = 112 - 112
    x² + 6x - 112 = 0

    By factoring, x² + 6x - 112 = (x - 8)(x + 14). So, the previous equation           becomes

    (x - 8)(x +14) = 0

   So, either 

    x - 8 = 0 or x + 14 = 0

   Thus, x = 8 or x = -14. However, since x represents the length of the base and the length is always positive, it cannot be negative. Hence, x = 8. Therefore, the length of the base is 8 cm.

2. Let x = length of increase in both length and width of the rectangular garden

Then,

14 + x = length of the new rectangular garden
12 + x = width of the new rectangular garden

So, 

(Area of the new garden) = (length of the new garden)(width of the new garden) 

255 = (14 + x)(12 + x) (1)

Note that 

(14 + x)(12 + x) = (x + 14)(x + 12)
                          = x(x + 14) + 12(x + 14)
                          = x² + 14x + 12x + 168 
                          = x² + 26x + 168

So, the equation (1) becomes

255 = x² + 26x + 168

By symmetric property of equations, we can interchange the side of the previous equation so that 

x² + 26x + 168 = 255

To make the right side becomes 0, we subtract both sides by 255:

x² + 26x + 168 - 255 = 255 - 255
x² + 26x - 87 = 0 

To solve the preceding equation, we use the quadratic formula.

First, we let

a = numerical coefficient of x² = 1

Note: if the numerical coefficient is hidden, it is automatically = 1.

b = numerical coefficient of x = 26
c = constant term = - 87

Then, using the quadratic formula 

x =  \frac{-b \pm  \sqrt{b^2 - 4ac} }{2a} =  \frac{-26 \pm  \sqrt{26^2 - 4(1)(-87)} }{2(1)}  &#10;\newline x =  \frac{-26 \pm  \sqrt{1,024} }{2}&#10;\newline&#10;\newline x =  \frac{-26 \pm  32 }{2}

So, 

x = \frac{-26 + 32 }{2} \text{  or } x = \frac{-26 - 32 }{2}&#10;\newline x = \frac{6 }{2} \text{  or } x = \frac{-58 }{2}&#10;\newline \boxed{ x = 3 \text{  or } x = -29}

Since x represents the amount of increase, x should be positive.

Hence x = 3.

Therefore, the length of the new garden is given by 

14 + x = 14 + 3 = 17 m.

3. The area of the shaded region is given by

(Area of shaded region) = π(outer radius)² - π(inner radius)²
                                       = π(2x)² - π6²
                                       = π(4x² - 36)

Since the area of the shaded region is 160π square centimeters,

π(4x² - 36) = 160π

Dividing both sides by π, we have 

4x² - 36 = 160

Note that this equation involves only x² and constants. In these types of equation we get rid of the constant term so that one side of the equation involves only x² so that we can solve the equation by getting the square root of both sides of the equation.

Adding both sides of the equation by 36, we have

4x² - 36 + 36 = 160 + 36
4x² = 196 

Then, we divide both sides by 4 so that

x² = 49

Taking the square root of both sides, we have

x = \pm 7

Note: If we take the square root of both sides, we need to add the plus minus sign (\pm) because equations involving x² always have 2 solutions.

So, x = 7 or x = -7.

But, x cannot be -7 because 2x represents the length of the outer radius and so x should be positive.

Hence x = 7 cm

4. At time t, h(t) represents the height of the object when it hits the ground. When the object hits the ground, its height is 0. So,
 
h(t) = 0   (1)

Moreover, since v_0 = 27 and h_0 = 10, 

h(t) = -16t² + 27t + 10   (2)

Since the right side of the equations (1) and (2) are both equal to h(t), we can have

-16t² + 27t + 10 = 0

To solve this equation, we'll use the quadratic formula.

Note: If the right side of a quadratic equation is hard to factor into binomials, it is practical to solve the equation by quadratic formula. 

First, we let

a = numerical coefficient of t² = -16 
b = numerical coefficient of t = 27
c = constant term = 10

Then, using the quadratic formula 

t = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-27 \pm \sqrt{27^2 - 4(-16)(10)} }{2(-16)} \newline t = \frac{-27 \pm \sqrt{1,369} }{-32} \newline \newline t = \frac{-27 \pm 37 }{32}

So, 

t = \frac{-27 + 37 }{-32} \text{ or } t = \frac{-27 - 37 }{-32} \newline t = \frac{-10}{32}  \text{ or } t = \frac{-64 }{-32}   \newline \boxed{ t = -0.3125 \text{ or } t = 2}

Since t represents the amount of time, t should be positive. 

Hence t = 2. Therefore, it takes 2 seconds for the object to hit the ground.


 




 





3 0
3 years ago
Read 2 more answers
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