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Strike441 [17]
2 years ago
13

Two consecutive even integers have the following property. When the smaller integer is added to three times the larger integer,

the result is two less than five times the smaller integer.
Rafael tries to model this scenario with the equation shown below. Unfortunately, Rafael has made an error on the left side of the equation. Explain what error he made.
n + 3n + 2 = 5n - 2
Write the correct equation (if you haven't already) and solve it to find the two consecutive even integers.
Mathematics
1 answer:
Valentin [98]2 years ago
6 0

Answer:

See below.

Step-by-step explanation:

Let the smaller integer be n.

Then the larger integer is n + 2.

"When the smaller integer is added to three times the larger integer, the result is two less than five times the smaller integer."

The correct equation is:

n + 3(n + 2) = 5n - 2

His mistake is not using parentheses on the left side. Since he left out the parentheses, the left side is not "the smaller integer added to three times the larger integer", but rather it is "the smaller integer added to 2 plus 3 times the smaller integer."

n + 3(n + 2) = 5n - 2

n + 3n + 6 = 5n - 2

4n + 6 = 5n - 2

8 = n

n = 8

n + 2 = 10

The integers are 8 and 10.

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A variable contains five categories. It is expected that data are uniformly distributed across these five categories. To test th
lubasha [3.4K]

Answer:

33.293 ± 0.01= 33.303 and 33.383,

Step-by-step explanation:

We first need to fit a normal distribution , but neither the mean  nor the standard deviation is given . We therefore estimate the sample mean and sample standard deviation  <em>s</em>. Using the data we find ∑fx=<u>378 </u>  and

<u>∑fx²=1344   </u> so that mean x` = 2.885 or 2.9   and standard deviation s =1.360

x        f                fx       x²         fx²

1       27             27        1          27

2       30           60         4          120

3       29           87         9          261

4       21            84         16        336

<u>5       24           120       25        600           </u>

<u>      ∑f=131      ∑fx=378             ∑fx²=1344   </u>

Mean = x`=<u> </u> ∑fx/ <u>  </u>∑f=  2.9

Standard Deviation = s= √∑fx²/∑f-(∑fx/∑f)²

                  s= √1344/131 - (378/131)²

                   s= √10.26-(2.9)²

                     s= √10.26- 8.41

                    s= √1.85= 1.360

Next we need to compute the expected frequencies for all classes and the value of chi square. The necessary calculations for expected frequencies , ei`s ( ei= npi`) where pi` is the estimate of pi together with the value of chi square are shown below.

Categories      zi`      P(Z<z)             pi`       Expected         Observed

                                                                    frequency ei    Frequency Oi

1                     -1.39      0.0823    0.0823       10.78                  27

2                   -0.66       0.2546     0.1723         22.57                30

3                    0.07        0.5279      0.2733       35.80                 29

4                   0.808      0.7881       0.2602        34.08                21

5                    1.54        0.937          0.1489        19.51                 24

Next we need to compute the expected frequencies for all classes and the value of chi square. The necessary calculations for expected frequencies , ei`s ( ei= npi`) where pi` is the estimate of pi together with the value of chi square are shown below.

Categories           Expected         Observed       (oi-ei)²/ei

                       frequency ei    Frequency Oi         OBSERVED VALUE

1                            10.78                  27                     24.41

2                          22.57                30                         1.54

3                          35.80                 29                        1.29

4                         34.08                21                           5.02

5                          19.51                 24                         1.033

<u>Total                                              131                   33.293</u>

There are five categories , we have used the sample mean and sample standard deviation , so the number of degrees of freedom is 5-1-2= 2

The critical region is chi square ≥ chi square (0.001)(2) =9.21

<u>CONCLUSION:</u>

Since the calculated value of chi square =9.21  does not fall in the critical region we are unable to reject our null hypothesis and conclude normal distribution provides a good fit for the given frequency distribution.

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Using the substitution P=1/x and Q=1/y.
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Step-by-step explanation:

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Answer:

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