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levacccp [35]
3 years ago
11

Can you help solve this

Mathematics
1 answer:
scoundrel [369]3 years ago
6 0
I believe the answer is b
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1)the middle value thats lies halfway along the sequence is the
svetoff [14.1K]

Answer:

1) C

2) A

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
2/3 + 3/4 + 4/2. If a fraction appears in your answer, it should be in simplified form.
V125BC [204]
We must first make each fraction have a common denominator, this will be 12
2/3 * 4/4 = 8/12
3/4 * 3/3 = 9/12
4/2 * 6/6 = 24/12
we can then add the numerators of each fraction
8 + 9 + 24 = 41
so our answer is 41/12
we can not simplify any further and thus get
41/12
5 0
3 years ago
The Wall Street Journal Corporate Perceptions Study 2011 surveyed readers and asked how each rated the Quality of Management and
kykrilka [37]

Answer:

The calculated χ² =  <u>17.0281</u>    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent

The p-value is 0 .001909. The result is significant at p < 0.05

Part b:

40 > 8.5

35> 7.5

25> 4

Step-by-step explanation:

1) Let the null and alternative hypothesis as

H0: the quality of management and the reputation of the company are independent

against the claim

Ha: the quality of management and the reputation of the company are dependent

2) The significance level alpha is set at 0.05

3) The test statistic under H0 is

χ²= ∑ (o - e)²/ e where O is the observed and e is the expected frequency

which has an approximate chi square distribution with ( 3-1) (3-1)=  4 d.f

4) Computations:

Under H0 ,

Observed       Expected E              χ²= ∑(O-e)²/e

40                      35.00                          0.71

25                      24.50                         0.01

5                         10.50                         2.88  

35                      40.00                         0.62

35                      28.0                          1.75

10                       12.00                           0.33  

25                      25.00                             0.00

10                        17.50                              3.21

<u>15                       7.50                                 7.50  </u>

<u>∑                                                               17.0281</u>

     

     

Column Totals 100 70 30   200  (Grand Total)

5) The critical region is χ² ≥ χ² (0.05)2 = 9.49

6) Conclusion:

The calculated χ² =  <u>17.0281</u>    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent.

7) The p-value is 0 .001909. The result is significant at p < 0.05

The p- values tells that the variables are dependent.

Part b:

If we take the excellent row total = 70 and compare it with the excellent column total= 100

If we take the good row total = 70 and compare it with the good column total= 80

If we take the fair row total = 50 and compare it with the fair column total= 30

The two attributes are said to be associated if

Thus we see that ( where (A)(B) are row and columns totals and AB are the cell contents)

AB> (A)(B)/N  

40 > 1700/200

40 > 8.5

35> 1500/200

35> 7.5

25> 800/200

25> 4

and so on.

Hence they are positively associated

8 0
3 years ago
Explain how the value of the 7 in 327,902 will change if you move it to the tens place
rjkz [21]
It will change from having a value of 7,000 to having a value of 70
6 0
3 years ago
Mark studies a group of 30 whales. If each whale weighed approximately 3.8 x 105 pounds, find the total weight of all 30 whales.
Yakvenalex [24]

Answer:

= 1.14 \times 10^{7}pounds

Step-by-step explanation:

If each whales weighed

3.8 \times  {10}^{5}

Then 30 whales weighed

30 \times 3.8 \times  {10}^{5}

=114 \times {10}^{5}

= 1.14 \times 10^{2}  \times {10}^{10}

Applying the law of indices we obtain

= 1.14 \times 10^{(2+ 5)}

=1.14 \times 10^{7}pounds

6 0
3 years ago
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