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igomit [66]
3 years ago
11

Two trucks are driving to the same place. The first truck starts 50 miles ahead of the second truck and travels at an average sp

eed of 60 miles per hour. The second truck travels at an average speed of 70 miles per hour. Which graph represents this situation and shows the number of hours it will take for the second truck to pass the first truck?
Mathematics
1 answer:
Snowcat [4.5K]3 years ago
3 0
This is a DRT (Distance, Rate, Time) problem. So, we can set up a table to get our answer. Our table should look something like this:
<u>                    </u><u>D         R        T</u>
<u>Slow Truck  d-50   60      t</u>
<u>Fast Truck   d         70      t</u>
Now that we have set up our table, we can use the values to find the answer. Starting off, we can use the rule T = D/R to get the equation \frac{d-50}{60} =  \frac{d}{70}. Now, we can simply solve for d, which can be done by cross multiplying and simplifying. After we do that, we can get that d = 350. Then, we can use the formula T = D/R, as stated previously, to substitute the rate and distance, which gives us t = 350/70. Simplifying that, we can get t = 5. Therefore, the time it takes for the second truck to pass the first truck is 5 hours. Hope this helped and have a fabulous day!
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Which product is positive?
Natalija [7]

Option D:

\left(-\frac{2}{5}\right)\left(-\frac{8}{9}\right)\left(\frac{1}{3}\right)\left(\frac{2}{7}\right) is positive product.

Solution:

<em>If the negative sign is in even number of times then the product is positive.</em>

<em>If the negative sign is in odd number of times then the product is negative.</em>

To find which product is positive:

Option A:

$\left(\frac{2}{5}\right)\left(-\frac{8}{9}\right)\left(-\frac{1}{3}\right)\left(-\frac{2}{7}\right)

Here, number of negative signs = 3

3 is an odd number.

So, the product is negative.

Option B:

$\left(-\frac{2}{5}\right)\left(\frac{8}{9}\right)\left(-\frac{1}{3}\right)\left(-\frac{2}{7}\right)

Here, number of negative signs = 3

3 is an odd number.

So, the product is negative.

Option C:

$\left(\frac{2}{5}\right)\left(\frac{8}{9}\right)\left(\frac{1}{3}\right)\left(-\frac{2}{7}\right)

Here, number of negative sign = 1

1 is an odd number.

So, the product is negative.

Option D:

$\left(-\frac{2}{5}\right)\left(-\frac{8}{9}\right)\left(\frac{1}{3}\right)\left(\frac{2}{7}\right)

Here, number of negative sign = 2

2 is an odd number.

So, the product is positive.

Hence option D is the correct answer.

\left(-\frac{2}{5}\right)\left(-\frac{8}{9}\right)\left(\frac{1}{3}\right)\left(\frac{2}{7}\right) is positive product.

4 0
3 years ago
Need answer fast as possible
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Answer:15

Step-by-step explanation:

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Answer:

top:correct

middle:incorrect

bottom:correct

Step-by-step explanation:

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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

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Step-by-step explanation:

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