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tatuchka
2 months ago
8

Consider the equation StartFraction x over 5 EndFraction minus 2 = 11, where x represents the number of basketball players that

have signed up for a new league. What is a possible scenario that would be modeled by this equation? Consider how many teams in the league and the number of players on each team this equation could describe?
Mathematics
2 answers:
tester [12.3K]2 months ago
5 0
This equation could depict a scenario that calculates how many players are allocated per team in a league consisting of five teams, considering that 2 players must be designated as referees or scorekeepers. Each team would then have 11 players.
lawyer [12.5K]2 months ago
4 1
Envision the equation StartFraction x over 5 EndFraction minus 2 = 11, where x denotes the total number of basketball players registered for a new league. What scenario might this equation represent? Think about the number of teams in the league and the players assigned to each team.
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A person flipped a coin 100 times and obtained 73 heads. Can the person conclude that the coin was unbalanced?
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Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0
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A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of awriting
tester [12383]

Answer:

a) Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

b) p_v =P(t_{17}

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c) p_v =P(t_{17}

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d) p_v =P(t_{17}

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

\bar X represents the average of the samples

s denotes the standard deviation of the samples

n=18 indicates the number of samples

\mu_o =10 is the value we are examining

\alpha defines the significance level for the test.

t represents the specific statistic of interest

p_v indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)

Part b

In this scenario t=-2.3, \alpha=0.05

Initially, we need to calculate the degrees of freedom df=n-1=18-1=17

Since this is a left-tailed test, the p-value is determined by:

p_v =P(t_{17}

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8, \alpha=0.01

We need to find the degrees of freedom df=n-1=18-1=17

For the left-tailed test, the p-value is given by:

p_v =P(t_{17}

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6, \alpha=0.05

Firstly, we find the degrees of freedom df=n-1=18-1=17

Since we are conducting a left-tailed test, the p-value is calculated as:

p_v =P(t_{17}

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

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2 months ago
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