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Rom4ik [11]
3 years ago
15

Ony scored 22 and 25 points in the first two games of a basketball tournament. Tony's goal is to make the All-Tournament Team. S

o far, in all of the tournaments he's been to, all of the players who made the All-Tournament Team scored an average between 27 and 32 points in the tournaments. Determine the range of points Tony should score in his next game to make the All-Tournament Team.
Mathematics
1 answer:
Westkost [7]3 years ago
6 0

Answer:

34 and 49

Step-by-step explanation:

Calculation to Determine the range of points that should be score in his next game

Based on the information given we were told that he scored 22 and 25 points in the first two games which means that his current scores will be : 22 and 25 point

Secondly we were told that Tony All-Tournament Team as well scored an average between 27 and 32 points in the tournaments based on this we are going to calculate Tony's next score in the next game.

Hence,

Let x be Tony's next score in the next game.

Now let calculate the lower required score of 27 points

(x + 22 + 25)/(25-22) = 27

(x + 22 + 25)/3 = 27

(X+47)=3*27

Collect like terms

x = 81 - 47

x= 34

Let calculate the upper required score of 32 points

(x + 22 + 25)/(25-22) = 32

(x + 22 + 25)/3 = 32

(X+47)=3*32

Collect like terms

x = 96 - 47

x= 49

Therefore the range of points that should be score in his next game will be 34 and 49

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The answer is 28 years

At = A0 * e^(-k * t)
At = 12 g
A0 = 15 g
k = 7.9 × 10^-3 = 0.0079 
t = ?

12 = 15 * e^(-0.0079 * t)
12/15 = e^(-0.0079 * t)
0.8 = e^(-0.0079 * t)

Logarithm both sides (because ln(e) = 1:
ln(0.8) = ln(e^(-0.0079 * t))
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Finding the X-intercept.

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6 0
3 years ago
The distribution of heights for adult men in a certain population is approximately normal with mean 70 inches and standard devia
KengaRu [80]

Answer:

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12]

Step-by-step explanation:

I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.

In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1

If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)

W = \frac{X-\mu}{\sigma} = \frac{X-70}{4}

The values of the cumulative distribution function of W, denoted by \phi , can be found in the attached file. Lets call y = \frac{z-70}{4} . We have

0.1 = P(X > z) = P(\frac{X-70}{4} > \frac{z-70}{4}) = P(W > y) = 1-\phi(y)

Thus

\phi(y) = 1-0.1 = 0.9

by looking at the table, we find that y = 1.28, therefore

\frac{z-70}{4} = 1.28\\z = 1.28*4+70 = 75.12

The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12] .

Download pdf
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