a map has a scale of 2in: 6mi. if a two twons on a map is 10 in apart then how far apart are the real twons

Mathematics

1 answer:

Brilliant_brown [7]2 years ago

7 0

Answer: 30 miles

Work Shown:

2 in : 6 mi

5*2 in : 5*6 mi

10 in : 30 mi

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I REALLY NEED HELP

alex41 [277]

The area is 180ft I think. You can do (15*12)/2 = 90. Since it’s two triangles you then do 90*2. I hope I am right

4 0

3 years ago

Read 2 more answers

An article reported that for a sample of 52 kitchens with gas cooking appliances monitored during a one-week period, the sample

kolbaska11 [484]

Answer:

a) $654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

b) $n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189$

Step-by-step explanation:

Part a

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:

$df=n-1=52-1=51$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that $t_{\alpha/2}=2.01$

Replacing we got:

$654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

Part b

The margin of error is given by :

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

The desired margin of error is ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , and we use an estimator of the population variance the value of 175 replacing into formula (b) we got: