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Iteru [2.4K]
3 years ago
8

An acrobat is on a platform that is 25 feet in the air. She jumps down in the initial velocity of 4 feet/seconds . Write the qua

dratic function to represent the height h of the acrobat t seconds after the jump. If a safety nets place 5 feet above the ground how long will it take for her to land safely on the net ?
Mathematics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

a) h(t) = -16t²+ 4t + 25

b) Therefore it would take her 1.25 seconds to land safely in the net.

Step-by-step explanation:

The equation to be used is given as

h(t) = –gt²+ vt + S(t)

where g = force of gravity and because our question is in feet's = 16

V= Initial velocity

h = height

S = distance

a) The quadratic function for the above question

Using the formula

v = Initial velocity in the question = 4ft/s

S(t) = 25ft in the air

Hence the Quadratic function is

h(t) = –gt²+ vt + S(t)

h(t) = -16t²+ 4t + 25

b) If the safety net is placed 5 feet above the ground, how long will it take her to land safely in the net?

We are to find the time(t)

We can calculate this by using the quadratic function we derived in question (a)

The quadratic function is

h(t) = -16t²+ 4t + 25

Where, the height (h) = 5 feet above the ground

5= -16t² + 4t + 25

-16t² + 4t + 25 -5 = 0

-16t² + 4t + 20 = 0

Using the quadratic equation formula of

x = -b ± √b² -4ac /2a

where the quadratic equation =

ax² + bx + c = 0

a = -16 , b = 4 , c = 20

x = -16 ± √4² - 4×-16×20 / 2×-16

x = -1 or ,1.25

Therefore it would take her 1.25 seconds to land safely in the net

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mrs_skeptik [129]

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Solution: Because we are given, each month the number of members gets triple, that means each month we can multiply by 3 to the number of members of previous month.

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7 0
3 years ago
A college infirmary conducted an experiment to determine the degree of relief provided by three cough remedies. Each cough remed
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Answer:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed. So we can say that the 3 remedies ar approximately equally effective.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     13                      9               33

Some relief               32                   28                     27              87

Total relief                7                       9                      14               30

Total                         50                    50                    50              150

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

The level os significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*33}{150}=11

E_{2} =\frac{50*33}{150}=11

E_{3} =\frac{50*33}{150}=11

E_{4} =\frac{50*87}{150}=29

E_{5} =\frac{50*87}{150}=29

E_{6} =\frac{50*87}{150}=29

E_{7} =\frac{50*30}{150}=10

E_{8} =\frac{50*30}{150}=10

E_{9} =\frac{50*30}{150}=10

And the expected values are given by:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     11                       11               33

Some relief               29                   29                     29              87

Total relief                10                     10                     10               30

Total                         50                    50                    50              150

And now we can calculate the statistic:

\chi^2 = \frac{(11-11)^2}{11}+\frac{(13-11)^2}{11}+\frac{(9-11)^2}{11}+\frac{(32-29)^2}{29}+\frac{(28-29)^2}{29}+\frac{(27-29)^2}{29}+\frac{(7-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(14-10)^2}{10} =3.81

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(3-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.81,4,TRUE)"

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.

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3 years ago
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Answer:

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Step-by-step explanation:

t = total # of marbles

Equation: t = 4 x 12

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2 years ago
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