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kirill115 [55]
4 years ago
6

There are 60 treats. he is allowed 2 treats a day. how many treats will he have left after 10 days?

Mathematics
1 answer:
nydimaria [60]4 years ago
6 0
2 x 10=20 60-20=40. 40 treats left
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The length of a parasite in experiment A is 2.5 × 10–3 inch. The length of a parasite in experiment B is 1.25 × 10–4 inch. How m
torisob [31]

We have been given that :-

The length of a parasite in experiment A is 2.5 \times 10^{-3}

The length of a parasite in experiment B is 1.25 \times 10^{-4}

Let us write the the  length of the parasite in experiment A in the exponent of -3.

1.25 \times 10^{-4}= 0.125 \times 10^{-3}

Clearly, the length of parasite in experiment A is greater than the length of parasite in experiment B.

The difference in the length is given by

2.5 \times 10^{-3} -0.125 \times 10^{-4}

=2.375 \times 10^{-3}

Therefore, the length of the parasite in experiment A is =2.375 \times 10^{-3} inches  greater than the length of the parasite in experiment B.

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Square root of 1269 = 35.62
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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

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3 years ago
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Answer:

x= number of tickets bought

y= total

44x+12=y

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They are both going 14 meters per second so therefor they will be the same distance apart the whole time.
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