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svetlana [45]
3 years ago
5

Which expression is equivalent to (-7m* + 15m2 – 12m?) + (3m® – 12m* -zma),

Mathematics
1 answer:
Leokris [45]3 years ago
8 0

Answer:

3m^4-m^3+2m^2

Step-by-step explanation:

Given the expression:

\frac{(-7m^4 + 15m^3-12m^2)+(3m^5-12m^4-7m^3)}{m-6} \\\\Firstly\ simplify\ the\ bracket\ should\ be\ :\\\\=\frac{-7m^4 + 15m^3-12m^2+3m^5-12m^4-7m^3}{m-6}\\\\collect\ like\ terms\ together:\\\\=\frac{3m^5-7m^4-12m^4 + 15m^3-7m^3-12m^2}{m-6}\\\\=\frac{3m^5-19m^4 + 8m^3-12m^2}{m-6}\\\\=\frac{m^2(3m^3-19m^2+8m-12)}{m-6}\\\\=\frac{m^2(m-6)(3m^2-m+2)}{m-6}  \\\\=m^2(3m^2-m+2)\\\\=3m^4-m^3+2m^2

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Irina18 [472]

The test statistics is used to determine if the anxiety medication can happen under a null hypothesis

The test statistics that is an appropriate hypothesis test is z =\frac{0.35 - 0.40}{\sqrt{\frac{0.35(1 - 0.35)}{60} + \frac{0.40(1 - 0.40)}{85}}}

<h3>How to determine the test statistic</h3>

<u>40 milligrams of medication</u>

Patients = 60

Lower stress level patients = 21

The mean of this medication is:

\bar x = \frac{x}{n}

So, we have:

\bar x_1 = \frac{21}{60}

\bar x_1 = 0.35

<u>75 milligrams of medication</u>

Patients = 85

Lower stress level patients = 34

The mean of this medication is:

\bar x = \frac{x}{n}

So, we have:

\bar x_2 = \frac{34}{85}

\bar x_2 =0.4

The test statistic is then calculated as:

z =\frac{\bar x_1 - \bar x_2}{\sqrt{\frac{\bar x_1(1 - \bar x_1)}{n_1} + \frac{\bar x_2(1 - \bar x_2)}{n_2}}}

The equation becomes

z =\frac{0.35 - 0.40}{\sqrt{\frac{0.35(1 - 0.35)}{60} + \frac{0.40(1 - 0.40)}{85}}}

Hence, the test statistics that is an appropriate hypothesis test is z =\frac{0.35 - 0.40}{\sqrt{\frac{0.35(1 - 0.35)}{60} + \frac{0.40(1 - 0.40)}{85}}}

Read more about test statistic at:

brainly.com/question/15980493

7 0
2 years ago
Factorise<br>x2 + 2x - 3​
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Answer:

( x − 1 ) ( x +3 )

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4 years ago
If x = 2 and y=5, what is the value of the expression below?
tangare [24]

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73, 81, 89, 75, 89, 86, 84, 78, 91, 78. What is the MAD
ruslelena [56]
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Miles draws three cards at random from a standard deck of 52 cards, without replacement.
o-na [289]

Answer:

P( red, red, red no replacement ) =  2/17

P( 3 cards with same rank, no replacement) =  1/425

Step-by-step explanation:

There are 52 cards, 26 are red

P( red) =  red/total =26/ 52 = 1/2

Now draw the second card without replacement

There are 51 cards, 25 are red

P( red) =  red/total =25/51

Now draw the third card without replacement

There are 50 cards, 24 are red

P( red) =  red/total =24/50 = 12/25

P( red, red, red no replacement ) = 1/2 * 25/51 * 12/25 = 2/17

P ( all have the same rank)

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We don't care what the first rank is, now the second and third draw have to have the same rank as the first

P( rank) = 1/1

Now draw the second card without replacement

There are 51 cards, 3 are left with the same rank

P( same rank) = cards with same rank/total = 3/51

Now draw the third card without replacement

There are 50 cards, 2 are left with the same rank

P( same rank) = cards with same rank/total = 2/50 = 1/25

P( 3 cards with same rank, no replacement) = 1 * 3/51*1/25 = 1/425

4 0
3 years ago
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