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emmasim [6.3K]
3 years ago
11

How to solve this question????? Help me please

Mathematics
1 answer:
julsineya [31]3 years ago
7 0

<em>Answer: 24/10</em>

<em>Answer: 24/10Step-by-step explanation:</em>

<em>1</em><em>)</em><em> </em><em>1*1÷10 +1*1÷5+2÷20</em>

<em>2</em><em>)</em><em>(1*10+1)/10+(1*5+1)/5+2/20 </em><em>3</em><em>)</em><em>11/10+6/5+2/20</em>

<em>4</em><em>)</em><em>11/10+6/5+1/10</em>

<em>5</em><em>)</em><em>11/10+12/10+1/10</em>

<em>6</em><em>)</em><em>(11+12+1)/10</em>

<em>7</em><em>)</em><em>24/10</em>

<em>Hope it helps!Please mark me brainest and follow me to get the answer faster</em>

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Answer:

-12a^{4}  + 3a^{3} + 20a^{2} -5a

Step-by-step explanation:

(5a-3a^{3} ) • (4a-1)

Let's use FOIL (first, outer, inner, last) to solve this. We'll multiply the terms by one another in that fashion.

20a^{2} - 5a - 12a^{4} + 3a^{3}

Rearrange in decreasing exponents.

-12a^{4}  + 3a^{3} + 20a^{2} -5a

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Complete Question

Answer:

a

  SE  = 0.66}

b

-3.29 <  \mu_1 - \mu_2 <  -0.70  

Step-by-step explanation:

From the question we are told that

  The sample size is  n  = 60

   The first sample mean is  \= x _1  =  8

    The second sample mean is   \= x _2  =  10

    The first variance is  v_1 =  0.25

    The first variance is  v_2 =  0.55

Given that  the confidence level is 95% then the level of significance is 5% =  0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the first standard deviation is  

     \sigma_1 =  \sqrt{v_1}

=>   \sigma_1 =  \sqrt{0.25}

=>   \sigma_1 =  0.5

Generally the second standard deviation is

     \sigma_2 =  \sqrt{v_2}

=>   \sigma_2 =  \sqrt{0.55}

=>   \sigma_2 =  0.742    

Generally the first standard error is

     SE_1  =  \frac{\sigma_1}{\sqrt{n} }

      SE_1  =  \frac{0.5}{\sqrt{60} }

     SE_1  =  0.06

Generally the second standard error is

     SE_2  =  \frac{\sigma_2}{\sqrt{n} }

      SE_2  =  \frac{0.742}{\sqrt{60} }

     SE_2  =  0.09

Generally the standard error of the difference between their mean scores is mathematically represented as    

      SE  =  \sqrt{SE_1^2 + SE_2^2 }

=>     SE  =  \sqrt{0.06^2 +0.09^2 }

=>     SE  = 0.66}

Generally 95% confidence interval is mathematically represented as  

      (\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } *  SE) <  \mu_1 - \mu_2 <  (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } *  SE)

=> (8 -10) -(1.96 *  0.66) <  \mu_1 - \mu_2 <  (8-10) +(Z_{\frac{\alpha }{2} } *  0.66)  

=>  -3.29 <  \mu_1 - \mu_2 <  -0.70  

 

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