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Tanzania [10]
3 years ago
15

If a body moves in a straight line according to the law s = 24t + 3t^2 - t^3, where s is the distance measured in meters from th

e origin and t is the time in seconds after it starts to move, calculate the body's velocity as a function of time.
A. 63 m/s
B. 15 m/s
C. 27 m/s
D. 81 m/s
Mathematics
1 answer:
Anastasy [175]3 years ago
4 0
As the comments state, the velocity is the derivative of the position.

Therefore, the velocity as function of time is:

ds / dt = 24 + 6t - 3t^2.

That is a parabola whose maximum is (1,27). With that you know that the velocity will never be either 63 m/s or 81 m/s.

Also, you know that the velocity at t = 1 s is 27 m/s.

And, you can also find that the velocity at t = 3 is 15 m/s.

I am confident on that this analysis solves your question. Else, insert a comment.
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Step-by-step explanation:

y -4x = -1

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How do you Solve d+7/−3=4?
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Answer:

One solution was found :

                  d = 19/3 = 6.333

Step-by-step explanation:

Step by step solution :

Step  1  :

            7

Simplify   ——

           -3

Equation at the end of step  1  :

        7    

 (d +  ——) -  4  = 0

       -3    

Step  2  :

Rewriting the whole as an Equivalent Fraction :

2.1   Adding a fraction to a whole

Rewrite the whole as a fraction using  -3  as the denominator :

         d     d • -3

    d =  —  =  ——————

         1       -3  

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

d • 3 + 7 • -1     3d - 7

——————————————  =  ——————

      3              3  

Equation at the end of step  2  :

 (3d - 7)    

 ———————— -  4  = 0

    3        

Step  3  :

Rewriting the whole as an Equivalent Fraction :

3.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  3  as the denominator :

        4     4 • 3

   4 =  —  =  —————

        1       3  

Adding fractions that have a common denominator :

3.2       Adding up the two equivalent fractions

(3d-7) - (4 • 3)     3d - 19

————————————————  =  ———————

       3                3  

Equation at the end of step  3  :

 3d - 19

 ———————  = 0

    3  

Step  4  :

When a fraction equals zero :

4.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

 3d-19

 ————— • 3 = 0 • 3

   3  

Now, on the left hand side, the  3  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

  3d-19  = 0

Solving a Single Variable Equation :

4.2      Solve  :    3d-19 = 0

Add  19  to both sides of the equation :

                     3d = 19

Divide both sides of the equation by 3:

                    d = 19/3 = 6.333

One solution was found :

                  d = 19/3 = 6.333

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Answer:

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c) The lowest score that would meet the colleges requirement would be decreased to 26.388.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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In this problem, we have that:

\mu = 21.5, \sigma = 4.7

a. Find the lowest test score that a student could get and still meet the colleges requirement.

This is the value of X when Z has a pvalue of 1 - 0.12 = 0.88. So it is X when Z = 1.175.

Z = \frac{X - \mu}{\sigma}

1.175 = \frac{X - 21.5}{4.7}

X - 21.5 = 1.175*4.7

X = 27.0225

The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b. If 1300 students are randomly selected, how many would be expected to have a test score that would meet the colleges requirement?

Top 12%, so 12% of them.

0.12*1300 = 156

156 would be expected to have a test score that would meet the colleges requirement

c. How does the answer to part (a) change if the college decided to accept the top 15% of all test scores?

It would decrease to the value of X when Z has a pvalue of 1-0.15 = 0.85. So X when Z = 1.04.

Z = \frac{X - \mu}{\sigma}

1.04 = \frac{X - 21.5}{4.7}

X - 21.5 = 1.04*4.7

X = 26.388

The lowest score that would meet the colleges requirement would be decreased to 26.388.

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