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3 years ago

Which answers I should have to answer

Mathematics

1 answer:

Nana76 [90]3 years ago

6 0

Given:

The table of point scored in 5 games.

After game 6, the mean number of points scored per game is 9.

To find:

The points scored by Maya in game 6.

Solution:

Let x be the points scored by Maya in game 6. Then sum of scores in all 6 games is:

$Sum=8+12+10+6+14+x$

$Sum=50+x$

We know that, the formula for mean is:

$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Total number of observations}}$

So, the mean number of points scored per game is:

$\text{Mean}=\dfrac{50+x}{6}$

It is given that the mean number of points scored per game is 9.

$\dfrac{50+x}{6}=9$

$50+x=9\times 6$

$x=54-50$

$x=4$

Therefore, the correct option is A.

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Stacy rolls a pair of six-sided fair dice.

insens350 [35]

Answer:

Pr(the sum of the numbers rolled is either a multiple of 3 or an even number)= $\frac{2}{3}$

Step-by-step explanation:

Let A be the event "sum of numbers is multiple of 3"

and B be the event "sum is an even number".

As our dice has six sides, so the sample space of two dices will be of 36 ordered pairs.

|sample space | = 36

Out of which 11 pairs have the sum multiple of 3 and 18 pairs having sum even.

So Pr(A)= $\frac{11}{36}$

and Pr(B)= $\frac{18}{36}$

and Pr(A∩B) = $\frac{5}{36}$ , as 5 pairs are common between A and B.

So now Pr(A or B)= Pr(A∪B)

= Pr(A)+Pr(B) - Pr(A∩B)

= $\frac{11}{36}$ + $\frac{18}{36}$ - $\frac{5}{36}$

= $\frac{24}{36}$

= $\frac{2}{3}$

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3 years ago

Can use the Pythagorean theorem to solve for b if c is 13 and a is 12

Greeley [361]

Answer:b=5

Step-by-step explanation:the formular for pythagoras theory is

C^2=a^2+b^2

13^2=12^2+b^2

169=144+b^2

Collect like terms

169-144=b^2

b^2=25

Take the the square root of both sides

b=square root of 25

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3 years ago

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed

algol [13]

Answer:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

$p_v = P(\chi^2_{5} >5.860)=0.32$

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number: 1, 2 , 3 , 4 , 5 ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are given:

$O_{1}=27$ $O_{2}=31$

$O_{3}=42$ $O_{4}=40$

$O_{5}=28$ $O_{6}=32$

The expected values are given by:

$E_{1} =\frac{1}{6}*200=33.33$ $E_{2} =\frac{1}{6}*200=33.33$

$E_{3} =\frac{1}{6}*200=33.33$ $E_{4} =\frac{1}{6}*200=33.33$

$E_{5} =\frac{1}{6}*200=33.33$ $E_{6} =\frac{1}{6}*200=33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

Now we can calculate the degrees of freedom for the statistic given by:

$df=Categories-1=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >5.860)=0.32$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

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Answer:

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