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Leya [2.2K]
4 years ago
7

A famous Premier-league footballer was quoted on Match of the Day to have a goal-scoring statistic of 0.36

Mathematics
1 answer:
stepan [7]4 years ago
7 0

Answer:

13a.\  The \ probability\ of \ scoring\ a\  goal\  in\  a\  game\  is\  0.36\\\\b. 3 \ games

14.

a. \ \£90\\b. \ 26 \ weeks\\c. \ \£4,860

15. This is not possible since p(yellow)=0.10 which is less than the stated 0.35

Step-by-step explanation:

13 a. -A goal-scoring statistic is the probability of a player scoring one goal in any given game played.

-A 0.36 goal scoring statistic means that the player has a 0.36 or 36% chance of scoring a goal in any game that he is involved in.

b. To determine the number of games it takes to score a whole goal, we divide the probability by 1 goal:

Games=\frac{1 \ game}{p(goal)}\\\\=\frac{1}{0.36}\\\\=2.778\approx 3

Hence, it takes approximately 3 games to score a full goal.

14.-The cost of a GCSE retake is £600 and attracts a 15% deposit

#The 15% equivalent in actual pounds is calculated by multiplying the percentage by the total cost as:

C_g=0.15\times 600\\\\=\£90

Hence, the 15%  deposit amount equals £90

b.#The student pays the balance in a £20 per week scheme,the total number of weeks is calculated as:

t=\frac{Balance}{Rate}\\\\=\frac{600-90}{20}\\\\=\frac{510}{20}\\\\=25.5\approx 26

Hence, it takes 26 weeks to clear the balance.

c. Given that 10% is the equivalent of £450

-We divide this amount by 10% to get the 100% equivalent

#We know that 10%=0.10

100\%=\frac{540}{0.1}\\\\=\£5400

#Alternatively, 100% divided by 10% is 10. Multiply this value by £540:

=540\times 10\\\\=\£5400

We subtract the discount amount for the per-discount price:

Cost=Total -discount\\\\=5400-540\\\\=\£4860

Hence, it will cost £4,860 without the discount.

15. Since we are not given the proportion of colors in the bag, we assume that all the 10 beads have different colors.

-As such, the sample space is 10 and each color has an equal chance of being picked:

P(Each \ Color)=P(Yellow)=\frac{1}{10}\\\\\therefore P(Yellow)

Hence, this is impossible since 0.10<0.35

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Answer:

  f^{(k)}(x)=\dfrac{17k!(-1)^k}{(x-9)^{k+1}}

Step-by-step explanation:

The question presumes you have access to a computer algebra system. The one I have access to provided the output in the attachment. The list at the bottom is the list of the first four derivatives of f(x).

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The derivatives alternate signs, so (-1)^k will be a factor.

The numerators start at 17 and increase by increasing factors: 2, 3, 4, indicating k! will be a factor.

The denominators have a degree that is k+1.

Putting these observations together, we can write an expression for the k-th derivative of f(x):

  \boxed{f^{(k)}(x)=\dfrac{17k!(-1)^k}{(x-9)^{k+1}}}

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Answer:

a) n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401  

And rounded up we have that n=2401

b) n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22  

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

Part a

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.02 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

Since we don't have prior information about the population proportion we can use ase estimator \hat p =0.5. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401  

And rounded up we have that n=2401

Part b

For thi case the estimator for p is \hat p =0.07

n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22  

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

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