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yawa3891 [41]
3 years ago
8

Egg measures 60 minus 26.3 in around the long axis how many inches in that

Mathematics
1 answer:
Sati [7]3 years ago
5 0
If you need anymore help let me know
You might be interested in
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
Volgvan

Answer: We reject the null hypothesis, and we use Normal distribution for the test.

Step-by-step explanation:

Since we have given that

We claim that

Null hypothesis : H_0:\mu\geq 50000

Alternate hypothesis : H_1:\mu

There  is 5% level of significance.

\bar{X}=46800\\\\\sigma=9800\\\\n=29

So, the test statistic would be

z=\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\z=\dfrac{46800-50000}{\dfrac{9800}{\sqrt{29}}}\\\\z=-1.75

Since alternate hypothesis is left tailed test.

So, p-value = P(z≤-2.31)=0.0401

And the P-value =0.0401 is less than the given level of significance i.e. 5% 0.05.

So, we reject the null hypothesis, and we use Normal distribution for the test.

4 0
3 years ago
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)}  
\newline x =  \frac{-26 \pm  \sqrt{1,024} }{2}
\newline
\newline x =  \frac{-26 \pm  32 }{2}

So, 

x = \frac{-26 + 32 }{2} \text{  or } x = \frac{-26 - 32 }{2}
\newline x = \frac{6 }{2} \text{  or } x = \frac{-58 }{2}
\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
4 years ago
Read 2 more answers
How do you determine if a given set of measurements works for the side lengths of a triangle? (It should be a number)
zaharov [31]
Add them all up and if they add up to 180 then your side lengths are good
5 0
4 years ago
Find x [Angles and Segment]
Alex_Xolod [135]

Answer:

  8.3 cm

Step-by-step explanation:

The product of lengths to the near and far point of intersection with the circle is the same in all cases:

  (7 cm)(7 cm) = (y)(11 cm +y) = (4 cm)(4 cm +x)

Since we're only interested in x, we can divide by 4 and subtract 4:

  49 cm² = (4 cm)(4 cm +x)

  (49/4) cm = 4 cm +x . . . . . . divide by 4 cm

  8.25 cm = x . . . . . . . . . . . . . subtract 4 cm

To the nearest tenth, x = 8.3 cm.

_____

For a tangent segment, the two points of intersection with the circle are the same point, so the product of lengths is the square of the length.

___

The angles depend on the size of the circle, which is not given.

6 0
3 years ago
A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the
amm1812

Answer:

s(t)=(ln|7|+ln|2|)\,ft

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\

Integrating both sides w.r.to t

v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\

Using initial value

v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}

We have to calculate the distance covered in time interval [0,5], so:

\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft

3 0
3 years ago
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