What is the slope of (2,2)

A map has a scale of 1:125 000. What length on the map represent an MRT Track that is 10 km long?

liubo4ka [24]

A scale of 1:125000 means that for any length that is on the map, the real life length is 125000 times as long. This also means that for any length in real life, the length on the map is 1/125000 times as long.

So, a real life length of 10km will be represented by a map length of 10km/125000 = 0.00008km = 8cm

8 0

4 years ago

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Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of bra

Stella [2.4K]

Answer:

The 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

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".

$\bar X_1 =114.1$ represent the sample mean 1

$\bar X_2 =129.9$ represent the sample mean 2

n1=5 represent the sample 1 size

n2=2 represent the sample 2 size

$s_1 =5.08$ sample standard deviation for sample 1

$s_2 =5.37$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =114.1-129.9=-15.8$

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

$df=n_1 +n_2 -1=5+5-2=8$

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(0.025,8)".And we see that $t_{\alpha/2}=\pm 2.31$

The standard error is given by the following formula:

$SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=3.306$

Confidence interval

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

$-15.8-2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-23.437$

$-15.8+2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-8.163$

So on this case the 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

8 0

3 years ago

Alfredo's painting his storage shed. He won't paint the top or bottom of the shed so he needs to calculate the lateral surface a

slavikrds [6]

Answer:

He is not painting the bottom or top meaning that he will paint 4 sides total, when there is a rectangular prism there are 6 sides and 3 pairs of 2 equal sides, so since there is a wall that is 9ft x 7 ft there is two of those, and since there is a wall that is 12 ft x 7 ft there are two of those as well. meaning that (7x9)·2=126 and (12x7)·2=168 when added we get 294. meaning that he needs to paint an area of 294 ft

7 0

2 years ago

Consider the line y=8x-2 find the equation of the line that is parallel to this line and passes through the point (-5,2)

DanielleElmas [232]

Answer:

the answer is y=8x+42

Step-by-step explanation:

y-y=m(x-x¹)

y-2=8(x-(-5))

y-2=8x+40

+2 +2

y=8x+42

8 0

3 years ago

Help with all of these porblems

777dan777 [17]

B² - 16a + 6

(7)² - 16(1/2) + 5
49 - 8 + 5
49 -13
36
-------------------------------
3b - 2a - 9
3(7) - 2(1/2) - 9
21 -1 - 9
20 - 9
11
-------------------------------
16a - b

16(1/2) - 7
8 - 7
1
-------------------------------
3(7) - 2(1/2) - 9
21 -1 - 9
20 - 9
11
-------------------------------
6(1/2)(7)
3(7)
21
-------------------------------

8 0

3 years ago