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3 years ago

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A soccer field measures 90 meters wide by 120 meters long starting together in one corner you walk the diagonal to the opposite

corner while your friend walks the length and with to get the same corner how much further dose your friend walk

1 answer:

disa [49]3 years ago

6 0

Answer:

The friend walks <em>60 m</em> further than the other person

Step-by-step explanation:

Given the dimensions of the soccer field:

Width = 90 m

Length = 120 m

One person walks along its diagonal.

As the length and width of the field are perpendicular, Diagonal can be calculated using Pythagorean Theorem:

$Diagonal ^2 =Length^2+Width^2$

$Diagonal ^2 =90^2+120^2\\\Rightarrow Diagonal ^2 =8100+14400\\\Rightarrow Diagonal ^2 =22500\\\Rightarrow Diagonal =150\ m$

So, the distance walked by the person along the diagonal = 150 m

Distance walked by the person's friend to reach the opposite point = Width + Length = 120 + 90 = 210 m

So, the difference of the two distances will give us the distance that was walked further by the friend.

The difference is = 150 - 90 = <em>60 m</em>

<em></em>

So, the friend walks 60 m further than the other person.

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If a new data point at 12 is added to the graph , which will be true ?

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Answer:

3. The mean will increase more than median, but both will increase.

Step-by-step explanation:

We have been given a dot plot. We are asked to find the true statement, if a new data point at 12 is added to our given dot plot.

Since we know that an large valued outlier affects mean more than median, so mean will increase more than median.

Let us check this using our given information. Our data set before adding 12 is:

1, 2, 2, 2, 3, 4, 4, 4, 5.

We can see that our data set has odd number (9) of data points, so median will be the value of 5th term, that is 3.

$\text{Mean}=\frac{1+2+2+2+3+4+4+4+5}{9}$

$\text{Mean}=\frac{27}{9}=3$

Therefore, mean of our data set is 3.

Upon adding a data point at 12, our new data set will be:

1, 2, 2, 2, 3, 4, 4, 4, 5, 12.

Now our data set has 10 terms, so median will be the average of 5th and 6th term.

$\text{Median}=\frac{3+4}{2}$

$\text{Median}=\frac{7}{2}=3.5$

Therefore, the median of new data set will be 3.5.

$\text{Mean}=\frac{1+2+2+2+3+4+4+4+5+12}{10}$

$\text{Mean}=\frac{39}{10}$

$\text{Mean}=3.9$

Therefore, mean of new data set will be 3.9.

Upon looking at our given statements we can see that 3rd statement is true as mean and median both increased after adding 12 to our data set, but mean increased more that median. Mean and median before adding 12 was 3; but after adding 12 median became 3.5, while mean became 3.9 (which is greater than mean).

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An example problem in a Statistics textbook asked to find the probability of dying when making a skydiving jump.

MArishka [77]

Answer:

(a) 0.999664

(b) 15052

Step-by-step explanation:

From the given data of recent years, there were about 3,000,000 skydiving jumps and 21 of them resulted in deaths.

So, the probability of death is $\frac{21}{3000000}==0.000007$ .

Assuming, this probability holds true for each skydiving and does not change in the present time.

So, as every skydiving is an independent event having a fixed probability of dying and there are only two possibilities, the diver will either die or survive, so, all skydiving can be regarded as is Bernoulli's trial.

Denoting the probability of dying in a single jump by $q$ .

$q=7\times 10^{-6}=0.000007$ .

So, the probability of survive, $p=1-q$

$\Rightarrow p=1-7\times 10^{-6}=0.999993.$

(a) The total number of jump he made, n=48

Using Bernoulli's equation, the probability of surviving in exactly 48 jumps (r=48) out of 48 jumps (n=48) is

$=\binom(n,r)p^rq^{n-r}$

$=\binom(48,48)(0.999993)^{48}(0.000007)^{48-48}$

$=(0.999993)^{48}=0.999664$ (approx)

So, the probability of survive in 48 skydiving is 0.999664,

(b) The given probability of surviving =90%=0.9

Let, total $n$ skydiving jumps required to meet the surviving probability of 0.9.

So, By using Bernoulli's equation,

$0.9=\binom {n }{r} p^rq^{n-r}$

Here, r=n.

$\Rightarrow 0.9=\binom{n}{n}p^nq^{n-n}$

$\Rightarrow 0.9=p^n$

$\Rightarrow 0.9=(0.999993)^n$

$\Rightarrow \ln(0.9)=n\ln(0.999993)$ [ taking $\log_e$ both sides]

$\Rightarrow n=\frac {\ln(0.9)}{\ln(0.999993)}$

$\Rightarrow n=15051.45$

The number of diving cant be a fractional value, so bound it to the upper integral value.

Hence, the total number of skydiving required to meet the 90% probability of surviving is 15052.

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