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Anna11 [10]
3 years ago
9

* Identify the first 6 multiples of 10​

Mathematics
2 answers:
bezimeni [28]3 years ago
8 0

Answer:

10, 20, 30, 40, 50, 60

Step-by-step explanation:

you just have to list the multiples

pls vote brainliest if this helped

telo118 [61]3 years ago
3 0

Answer:

10, 20, 30, 40, 50, 60

Step-by-step explanation:

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составь верные равенства и неравенства, используя таблицу: 32×2, 120+253, 900-653, 300×2, 640÷8, 900÷3, 450-350, 45×3
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A salesperson sold 30% of their books in stock on Monday. Then they sold 60% of their remaining books on Tuesday. What percent o
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Answer:

10%

Step-by-step explanation:

60% + 30%=90%

so they have 10%of the original amount left.

7 0
3 years ago
4 (3x + 2)-6<br> I neeeed help
yuradex [85]

Answer:

12x - 2

Step-by-step explanation:

Step 1: Write expression

4(3x + 2) - 6

Step 2: Distribute 4

12x + 8 - 6

Step 3: Combine like terms

12x - 2

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Read 2 more answers
Suppose you do not know the population mean fee charged to H&amp;R Block customers last year. Instead, suppose you take a sample
puteri [66]

Answer:

i \to a

    n = 96040000

i \to b

    n_1 =24010000

i \to c

    n_2 =41602500

ii\toa

     E = 58.16

ii\tob

    291.84  <  \mu  < 408.16\

ii\toc

    There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

     The sample size is n =  8

      The sample mean is  \= x  =  \$ 350    

      The sample standard deviation is  \$ 100

Considering question i

    i \to a

         At   E =  0.02  

given that the confidence level is 95%  =  0.95

         the level of significance would be  \alpha  =1-0.95 =  0.05

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

        Z_{\frac{ \alpha }{2} } =  1.96

So  the sample size is mathematically evaluated as

            n = [ \frac{Z_{\frac{\alpha }{2} } *  \sigma }{E} ]^2

=>        n =[ \frac{ 1.96 *  100}{ 0.02} ]^2

=>         n = 96040000

 i \to b

  At  E_1 = 0.04    and  confidence level  = 95%  =>  \alpha_1  = 0.05   =>  Z_{\frac{\alpha_1 }{2} } =  1.96

             n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } *  \sigma }{E_1} ]^2

=>           n_1 =[ \frac{ 1.96 *  100}{ 0.04} ]^2

=>           n_1 =24010000

 i \to c

       At   E_2 =  0.04     confidence level  = 99%  =>    \alpha_2  = 0.01

The critical value of  \frac{\alpha_2 }{2} from the normal distribution table is  

        Z_{\frac{ \alpha_2 }{2} } = 2.58

=>    n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } *  \sigma }{E_2} ]^2

=>    n_2 =[ \frac{ 2.58 *  100}{ 0.04} ]^2

=>    n_2 =41602500

Considering ii

Given that the level of significance is  \alpha  = 0.10

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

           Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as

          E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

substituting values

         E = 1.645  *  \frac{100 }{\sqrt{8} }

         E = 58.16

Generally the 90% confidence interval is mathematically evaluated as

         \= x - E <  \mu  <  \= x + E

=>      350 -  58.16  <  \mu  < 350 +  58.16

=>     291.84  <  \mu  < 408.16

So the interpretation is that there is 90% confidence that the mean  fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0
3 years ago
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