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3 years ago

I need help, I got 80 and 90 but it said I was wrong...

Mathematics

2 answers:

Anton [14]3 years ago

8 0

Answer:

I think it is 90 and 100

Step-by-step explanation:

I used this equation: x is the speed of the slower train

910 = (5x + 10) + 5x

Grace [21]3 years ago

5 0

Answer:

The answer is 86 and 96 mph

Step-by-step explanation:

Givens

Train A

Distance traveled = d

time taken = 5 hours

rate traveled = r

Train B

Distance traveled = 910 - d

time taken = 5 hours

rate = r + 10

Formula

basically d = r * t

Solution

Train A: d/r = 5 or d = 5*r

Train B = (910 - d)/(r + 10) = 5 or

Substitute d = 5*r into Train B

(910 - 5*r)/(r + 10) = 5 Multiply both sides by r + 10

(910 - 5r) = 5 * (r + 10) Remove the brackets on both sides

910 - 5r = 5r + 50 Add 5r to both sides

910 = 5r + 5r + 50

910 = 10r + 50 Subtract 50 from both sides

910 - 50 = 10r Combine the left

860 = 10r

10r = 860 Divide by 10

r = 860/10

r = 86

r + 10 = 96

Check

Train A distance: 5*r = 5 * 86 = 430

Train B distance: 5*r = 5 * 96 = 480

Total = 910

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3 years ago

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loris [4]

Answer:

Option D - $[0.12$

Step-by-step explanation:

Given : The manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be s=0.16.

To find : The 95% confidence interval for the population standard deviation sigma?

Solution :

Number of sample n=19

The degree of freedom is Df=n-1=19-1=18

The standard deviation of the sample is s=0.16

Applying chi-square table to find critical value,

Upper critical value of $\chi^2$ is $UC=\chi(\frac{0.05}{2},18) = 31.5264$

Lower critical value of $\chi^2$ is

$LC=\chi(1-\frac{0.05}{2},18) = 8.2307$

Lower limit of the 95% confidence interval for the population variance

$L=\frac{(df)\times (s^2)}{UC}$

$L=\frac{18\times (0.16^2)}{31.5264}$

$L=\frac{18\times0.0256}{31.5264}$

$L=\frac{0.4608}{31.5264}$

$L=0.0146$

Upper limit of the 95% confidence interval for the population variance

$U=\frac{(df)\times(s^2)}{LC}$

$U=\frac{18\times (0.16^2)}{8.2307}$

$U=\frac{18\times0.0256}{8.2307}$

$U=\frac{0.4608}{8.2307}$

$U=0.0559$

So, The 95% confidence interval for the population variance is [0.0146, 0.0560]

Now, The 95% confidence interval for the population standard deviation is

$[\sqrt{0.0146}$

$[0.1208$

or $[0.12$

Therefore, Option D is correct.

The 95% confidence interval for the population standard deviation is $[0.12$

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Answer:

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Step-by-step explanation:

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3 years ago

I need help, I got 80 and 90 but it said I was wrong...​

I need help, I got 80 and 90 but it said I was wrong...