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3 years ago

Can someone pls help me

Mathematics

1 answer:

nevsk [136]3 years ago

7 0

$\huge\text{Hey there!}$

$\huge\textsf{Your GUIDE}\huge\downarrow$

$\large\text{The DISTRIBUTIVE FORMULA is......}\\\large\text{a(b + c)}\\\large\text{a(b) + a(c)}}\\\large\text{a(b) = ab}\\\large\text{a(c)=ac}\\\large\text{= \underline{ab + ac}}$

$\text{Now lets SOLVE for your QUESTION, shall we?}}$

$\large\text{\bf{9(2n +1)}}\\ \large\text{\bf{9(2n) +9(1)}}\\\arge\text{\bf{9(2n)=18n}}\\\large\text{\bf{9(1)=9}}\\\large\text{\bf{= \underline{18n+9}}}}$

$\boxed{\boxed{\bold{Answer: 18n +9}}}\huge\checkmark$

$\large\textsf{Good luck on your assignment and enjoy your day!}$

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sdas [7]

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The total compound interest is $392.94.

Step-by-step explanation:

1,300 x 0.045=___ x 6

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4 years ago

Please help me answer this I'm having trouble

Alex777 [14]

The first one and the last one are correct.

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3 years ago

Read 2 more answers

What is cos 45 A. B. C. 1 D. E. F.

Leona [35]

I hope this helps you

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3 years ago

Assume that both populations are normally distributed. a) Test whether mu 1 not equals mu 2 at the alpha equals 0.05 level of s

Radda [10]

Answer:

$(16.2-14.1) -2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = -0.445$

$(16.2-14.1) +2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = 4.645$

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".

$n_1 = 19$ sample size 1

$n_2 = 19$ sample size 2

$\\bar X_1 = 16.2$ sample mean for group 1

$\\bar X_2 = 14.2$ sample mean for group 2

$s_1 = 4.5$ sample deviation for group 1

$s_2 = 3.1$ sample deviation for group 2

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

And the degrees of freedom are given by:

$df = n_1 +n_2 -2 = 19+19 -2= 36$

We want a 95% of confidence o then the significance level is 1-0.95 =0.05 and $\alpha/2 = 0.025$ if we find a critical value in the t distribution with 36 degrees of freedom we got:

$t_{cric} =2.03$

And replacing we got:

$(16.2-14.1) -2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = -0.445$

$(16.2-14.1) +2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = 4.645$

5 0

3 years ago

Four universities - 1, 2, 3, and 4 - are participating in a holiday basketball tournament. In first round, 1 will play 2 and 3 w

tia_tia [17]

Answer:

Part A. {S} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Part B. {A} = {1324, 1342, 1423. 1432}

Part C. {B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

Part D.

(A∪B) = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∩B) = ∅

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Step-by-step explanation:

A. All possible outcomes

There are four teams, each play a semi final where 1 and 2 plays against each other while 3 and 4 plays against each other. Winner of the first semi final can be either 1 or 2 therefore they both can not be in the championship game or in the losers game at the same time same goes for the other semi final.

Using this explanation (1324 denotes: 1 beats 2 and 3 beats 4 in first-round games and then 1 beats 3 and 2 beats 4), All possible outcomes are

{S} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

B. Event A in which 1 wins the tournament

From {S} we only have to write the outcomes in which 1 is the first number in 4digit combinations given in part A

{A} = {1324, 1342, 1423. 1432}

C. Event B in which 2 gets into championship game

From {S} we only have to write the outcomes in which 2 is either the first or second digit in 4digit combinations given in part A

{B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

D. Outcomes in (A∪B), (A∩B) and A'

I. (A∪B)

(A∪B) means A union B therefore all we have to do is combine all the members of A and B

(A∪B) = {1324, 1342, 1423, 1432} ∪{2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∪B) = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

II. (A∩B)

(A∩B) means A intersection B in which we have to find the common members of A and B. If there are no common members then the result of (A∩B) is a null set.

(A∩B) = {1324, 1342, 1423, 1432} ∩ {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∩B) = ∅

III. {A'}

A' means A compliment, in other words it can be described as all the possible outcomes that are not part of A. So all we do is to subtract outcomes of A from the total possible outcomes S

{A'} = {S} - {A}

{A'} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231} - {1324, 1342, 1423. 1432}

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

7 0

3 years ago