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Tatiana [17]
2 years ago
12

Kelly's basketball team made a total of 33 two points and three point baskets in their game last week. If her team scored a tota

l of 72 points how many two points baskets did they make
Mathematics
1 answer:
vladimir2022 [97]2 years ago
4 0

Answer:

They make <u>27</u> two points basket.

Step-by-step explanation:

Given:

Kelly's basketball team made a total of 33 two points and three point baskets in their game last week.

If her team scored a total of 72 points.

Now, to find the number of two points baskets they make.

Let the two points basket be x.

And the three points basket be y.

So, total number of two points and three point baskets are:

x+y=33

y=33-x\ \ \ .........(1)

Now, the total points team scored:

2x+3y=72

Substituting the value of y from equation (1) we get:

2x+3(33-x)=72

2x+99-3x=72

<em>Subtracting both sides by 99 we get:</em>

2x-3x=-27

-x=-27

<em>Dividing both sides by -1 we get:</em>

x=27.

The the two points basket = 27.

Therefore, they make 27 two points basket.

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Consider a sample with data values of 27, 25, 20, 15, 30, 34, 28, and 25. Compute the range, interquartile range, variance, and
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So we have a set of integers,

X=\{15,20,25,25,27,28,30,34\}

And we need to compute some stats.

First the range,

The range of data set is difference between the maximum and the minimum of the data set.

So the minimum is 15 and maximum is 34 therefore the range is 34 - 15 = 19.

Second the interquartile range,

The interquartile range of data set is the difference of the first and third quartiles.

First quartile is the value separating the lower quarter and higher three - quarters of the set. The first quartile is computed by taking median of the lower half of the sorted set. This is 15, 20, 25, 25 and it's median is 22.5

Third quartile is therefore the median of 27, 28, 30, 34 which is 29.

Next up, variance.

The variance of set measures how much data is spread out. For data set x_1,\dots, x_n with an average a,

\mathsf{var}(X)=\Sigma_{i=1}^{n}\dfrac{(x_i-a)^2}{n-1}

So first compute the average value which is actually a mean,

a=\dfrac{1}{2}\Sigma_{i=1}^{n}a_i

The sum \Sigma_{i=1}^{n}a_i of numbers in set is 204.

Divide this by number of elements in the set (8).

a=\dfrac{204}{8}=25.5

Then compute the variance,

\mathsf{var}(X)=\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}\approx\boxed{34.57}

And finally standard deviation,

Since have computed variance the standard deviation is nothing too hard. It's defined as a square root of variance,

\mathsf{sde}(X)=\sqrt{\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}}=\sqrt{34.57}\approx\boxed{5.88}

Hope this helps.

r3t40

6 0
2 years ago
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