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crimeas [40]
2 years ago
14

A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hit the ground. a.Find

the velocity of the object when it is 30.0 m above the ground. b.Find the total distance the object travels during the fall.
Mathematics
1 answer:
Lyrx [107]2 years ago
4 0
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2 
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
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We are given the following in the question:

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A sociologist wants to determine if the life expectancy of people in africa is less than the life expectancy of people in asia.
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Answer:

The sample statistics are attached.

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The alternative hypothesis should be: H_{a}: u_{africa}

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Replacing all the values into the formula, we have:

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Now, we using a level of significance of 0.01, and a degree of freedom (df) of 34, we use the t-table to find the p-value for this results. (df = N -1; in this case, we take the smaller sample, which is 35, giving us 34 of df).

Therefore, according to the table attached, <em>the p-value is less than 0.01,</em> which is less than our level of significance. This result means that the null hypothesis is rejected. In other words, there's enough evidence to say that <em>the life expectancy of people in Africa is less than the life of expectancy of people in Asia.</em>

<em></em>

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2 years ago
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The probabilities of the orphaned pets in six cities' animal shelters being different types of animals are given in the table. I
olga2289 [7]
Given the table below showing t<span>he probabilities of the orphaned pets in six cities' animal shelters being different types of animals.

</span>
   City          Cat          Dog        Dog        Dog       <span>Dog                                      
                         −Lhasa Apso −Mastiff −Chihuahua −Collie
Austin       24.5%      2.76%    2.86%     3.44%    2.65%
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Salt Lake
City          28.9%     2.85%     2.78%     2.96%    2.46%
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Total        22.9%     2.91%     2.68%     3.09%    2.58%</span><span>

Condtional probability </span><span>is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B).

The probability of A given B, is given by
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From the table, the probability that an </span>orphaned pet in a St. Louis animal shelter is a mastiff dog is 2.46% and the probability that <span>an <span>orphaned pet in a St. Louis animal shelter is a dog is 2.91% + 2.68% + 3.09% + 2.58% = 11.26%</span>

The probability that a randomly selected orphaned pet in a St. Louis animal shelter is a mastiff given that it is a dog, is given by
\frac{P(St. \, Louis\, and\, dog-mastiff)}{P(dog)} = \frac{2.46}{11.26} \times100\%=21.85\%</span>
5 0
2 years ago
Read 2 more answers
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