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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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Step-by-step explanation:

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Alisiya [41]

Answer:

$0.82-2.977\frac{0.048}{\sqrt{15}}=0.783$

$0.82+2.977\frac{0.048}{\sqrt{15}}=0.857$

So on this case the 99% confidence interval would be given by (0.783;0.857)

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X= 0.82$ represent the sample mean

$\mu$ population mean (variable of interest)

s=0.048 represent the sample standard deviation

n=15 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=15-1=14$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that $t_{\alpha/2}=2.977$

Now we have everything in order to replace into formula (1):

$0.82-2.977\frac{0.048}{\sqrt{15}}=0.783$

$0.82+2.977\frac{0.048}{\sqrt{15}}=0.857$

So on this case the 99% confidence interval would be given by (0.783;0.857)

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval

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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...