143 players per 11 teams.how many players are on each one

What is the distance between -8/5 and -3/5<br> A.-1<br> B.1<br> C.11/5<br> D. -11/5

Maurinko [17]

Answer:

1

Step-by-step explanation:

Subtract the two numbers and take the positive value ( since distance is positive)

-8/5 - (-3/5)

-8/5 + 3/5

-5/5

-1

Taking the positive value, 1

4 0

3 years ago

The average yield of the standard variety off soybeans per acre on a farm in a region is 48.8 bushels per acre, with a standard

I am Lyosha [343]

Answer:

$z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3$

$p_v =P(Z>2.3)=1-P(Z$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.

Step-by-step explanation:

1) Data given and notation

$\bar X=53.4$ represent the sample mean

$\sigma=12$ represent the population standard deviation assumed

$n=36$ sample size

$\mu_o =48.8$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is higher than 48.8, the system of hypothesis would be:

Null hypothesis: $\mu \leq 48$

Alternative hypothesis: $\mu > 48$

Since we assume that know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

3) Calculate the statistic

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

$z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3$

4)P-value

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

$p_v =P(Z>2.3)=1-P(Z$

5) Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.

3 0

3 years ago

Whats the slope on the graph?

oee [108]

Answer:

-9/5 is the slope

Step-by-step explanation:

Please give brainliest :)

6 0

2 years ago

What is the value of x?

alexgriva [62]

Answer:

x = 84

Step-by-step explanation:

If you add up all of the internal angles in ANY pentagon it will be 540, so

x = 540 - (131+108+107+110)

4 0

2 years ago

Read 2 more answers

A country's population in 1994 was 182 million. In 2002 it was 186 million. Estimate the population in 2004 using the exponentia

iogann1982 [59]

$\bf =ae^{kt}\qquad 
\begin{cases}
1994\impliedby \textit{year 0, starting point}\\
t=0\qquad P=182
\end{cases}\implies 182=ae^{k0}
\\\\\\
182=a\cdot e^0\implies 182=a\cdot 1\implies 182=a
\\\\\\
thus\qquad P=182e^{kt}\\\\
-------------------------------\\\\$

$\bf P=182e^{kt}\qquad 
\begin{cases}
2002\impliedby \textit{8 years later}\\
t=8\qquad P=186
\end{cases}\implies 186=182e^{k8}
\\\\\\
\cfrac{186}{182}=e^{8k}\implies ln\left( \frac{93}{91} \right)=ln(e^{8k})\implies ln\left( \frac{93}{91} \right)=8k
\\\\\\
\cfrac{ln\left( \frac{93}{91} \right)}{8}=k\implies 0.0027\approx k\implies \boxed{P=182e^{0.0027t}}$

what's the population in 2004? well, from 1994 to 2004 is 10 years later, so t = 10

plug that in, to get P for 2004

3 0

3 years ago