How to solve 2/3x-36>22

You coach a basketball ball team of 12 players; 5 players must be on the floor at all times; Figuring that every player can play

makvit [3.9K]

The function "choose k from n", nCk, is defined as
nCk = n!/(k!*(n-k)!) . . . . . where "!" indicates the factorial

a) No position sensitivity.
The number of possibilities is the number of ways you can choose 5 players from a roster of 12.
12C5 = 12*11*10*9*8/(5*4*3*2*1) = 792
You can put 792 different teams on the floor.

b) 1 of 2 centers, 2 of 5 guards, 2 of 5 forwards.
The number of possibilities is the product of the number of ways, for each position, you can choose the required number of players from those capable of playing the position.
(2C1)*(5C2)*(5C2) = 2*10*10 = 200
You can put 200 different teams on the floor.

4 0

3 years ago

Find the volume of a spere with radius of 7

choli [55]

Formula for volume of sphere V = 4/3 (Pi) r^3, where Pi ~ 3.14 and r is the radius = 7.

Therefore V = 4/3 * 3.14 * 7^3

= 4/3 * 3.14 * 343

= 4/3 * 1077.02

= 1,436.02667 cubic units

6 0

3 years ago

Read 2 more answers

Which angle is supplementary to 4?

Natalija [7]

2 !! hope it helped

8 0

3 years ago

Read 2 more answers

How many miles per hour can lorenzo ride his bike

Degger [83]

The answer is around 10.22 mph

5 0

3 years ago

Some research suggests that police officers are more likely to make an arrest in the presence of bystanders. If mentally disorde

zmey [24]

Answer:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

$p_v =P(t_{(19)}>1.638)=0.0589$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=7.03$ represent the sample mean

$s=9.42$ represent the sample standard deviation

$n=20$ sample size

$\mu_o =3.58$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 3.58 :

Null hypothesis: $\mu \leq 3.58$

Alternative hypothesis: $\mu > 3.58$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since is a right tailed test the p value would be:

$p_v =P(t_{(19)}>1.638)=0.0589$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

7 0

3 years ago