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Hoochie [10]
3 years ago
13

Suppose a researcher is testing to see if a basketball player can make free throws at a rate higher than the NBA average of 75%.

The player is tested by shooting 10 free throws and makes 8 of them. In conducting the related test of significance we have a computer applet do an appropriate simulation, with 1000 repetitions, and produce a null distribution. This distribution represents:_______.
A. Repeated results if the player makes 75% of his shots in the long run.
B. Repeated results if the player makes 80% of his shots in the long run.
C. Repeated results if the player makes more than 75% of his shots in the long run.
D. Repeated results if the player makes more than 80% of his shots in the long run.
E. None of these
Mathematics
1 answer:
stich3 [128]3 years ago
3 0

Answer: A. Repeated results if the player makes 75% of his shots in the long run.

Step-by-step explanation:

The null distribution is always the opposite of the alternative distribution which in most cases represents the claim or hypothesis which is to be tested or performed. In the scenario given, the challenge is to show that a basketball player has an average higher than that of the NBA. NBA average stands at 75%. The alternative hypothesis is the claim, which is ;

H1 : μ > 75%

THE null is thus :

H0 : μ = 75% ; which means that repeated result of the player will yields an average of 75%

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Answer:

3.5 square cm

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7 0
3 years ago
Click the picture that I sent you
Digiron [165]

Answer: 1211.6585 years

<u>Step-by-step explanation:</u>

The equation for exponential growth is: P=P_oe^{kt}

  • P: final population
  • P₀: initial population
  • k: rate of decay (or growth)
  • t: time

Use the half life information to find k:

\dfrac{1}{2}P_o=P_oe^{k(800)}\\\\\\\dfrac{1}{2}=e^{800k}\qquad \rightarrow \qquad \text{divided both sides by}\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=800k\qquad \rightarrow \qquad \text{applied ln to both sides}\\\\\\\dfrac{ln(\frac{1}{2})}{800}=k\qquad \rightarrow \qquad \text{divided both sides by 800}\\\\\\\large\boxed{-0.000867=k}

Next, input the information (65% decayed = 35% remaining) and the k-value to find your answer.

.35P_o=P_oe^{-0.000867t}\\\\\\.35=e^{-0.000867t}\\\\\\ln(.35)=-0.000867t\\\\\\\dfrac{ln(.35)}{-0.000867}=t\\\\\\\large\boxed{1211.6585=t}

5 0
3 years ago
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Formula = Population after n years = 2000( 1.04)^n

15 years the population will be 2000(1.04)^15

=3602.

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