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vovikov84 [41]
3 years ago
12

a party of trekkers left a staging hut, an elevation of 465 ft and arrived at their destination at an elevation of 2347 ft. acco

rding to the map, the camps were 3000 ft apart . what was average slope between the two camps
Mathematics
1 answer:
bazaltina [42]3 years ago
8 0
Ok, here we go.
The x-value is given(Great!), so now we need to find y-value for slope
\frac{2347-465}{3000}
\frac{1882}{3000}
The average slope, according to the calculator, is 0.627
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4y=x 3x-y=70 <br>how to solve this problem using substitution
diamong [38]
4y=x
3(4y)-y=7012y-y=70
11y=70
y=70/11
4(70/11)=x
280/11=x
3 0
4 years ago
Math Help. Greatly Appreciated
Triss [41]
So we know that the perimeter is 1260 feet and that the length is 100 more than the width, but we don't know that the width is.
So we're going to start with finding the width.
w = (x + x + 100) = 1260
from there we need to combine like terms
x + x = 2x
now we have
2x + 100  = 1260
From there, we will need to remove 100 from both sides, leaving us with
2x = 1160
now we'll need to divide both sides by 2
2x \div 2x = x
1160 \div 2 = 580
We now know that the width is 580, or more simply put, x=580.

Finding the length will be the easy part, as we know that it is just 100 more than the width
580 + 100 = 680
If you want an easier way of finding the length, just subtract the width from the perimeter.
1260 - 580 = 680
Leaving the final answer to be
Width = 580
Length = 680

3 0
3 years ago
Why doesnt Orgo eat Cabbage, Corn, Chicken, Clams, Cake, or Celery?
Vanyuwa [196]
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7 0
3 years ago
Real world inequalities:<br><br><br><br> You must be 4 feet or taller to ride the roller coaster
Anika [276]

\qquad\qquad\huge\underline{{\sf Answer}}

The required inequality that represent the given scenario is :

\qquad \tt \dashrightarrow \:x \geqslant 4

where, x is height in feet.

8 0
3 years ago
Suppose your statistics instructor gave six examinations during the semester. You received the following grades (percent correct
andriy [413]

Answer:

a. 15

b.

Sr.no Samples Sample mean

1           (79,64)         71.5

2          (79,84)          81.5

3          (79,82)          80.5

4          (79,92)          85.5

5          (79,77)           78

6         (64,84)            74

7          (64,82)            73

8          (64,92)            78

9          (64,77)            70.5

10        (84,82)            83

11         (84,92)            88

12        (84,77)             80.5

13        (82,92)            87

14        (82,77)            79.5

15        (92,77)            84.5

c.

mean of sample mean=population mean=79.67

Step-by-step explanation:

a.

The different samples of two test grade are nCr, where n=6 and r=2.

nCr=6C2=6!/2!(6-2)!=6*5*4!/2!4!=30/2=15.

Thus, there are 15 different samples of two test grade.

b.

All possible samples are listed below:

Sr.no Samples

1           (79,64)        

2          (79,84)          

3          (79,82)          

4          (79,92)        

5          (79,77)          

6         (64,84)            

7          (64,82)

8          (64,92)

9          (64,77)

10        (84,82)

11         (84,92)

12        (84,77)

13        (82,92)

14        (82,77)

15        (92,77)

The sample means for each sample can be calculated as

Sr.no Samples Sample mean

1           (79,64)         (79+64)/2=71.5

2          (79,84)          (79+84)/2=81.5

3          (79,82)          (79+82)/2=80.5

4          (79,92)          (79+92)/2=85.5

5          (79,77)           (79+77)/2=78

6         (64,84)            (64+84)/2=74

7          (64,82)            (64+82)/2=73

8          (64,92)            (64+92)/2=78

9          (64,77)            (64+77)/2=70.5

10        (84,82)            (84+82)/2=83

11         (84,92)            (84+92)/2=88

12        (84,77)             (84+77)/2=80.5

13        (82,92)            (82+92)/2=87

14        (82,77)            (82+77)/2=79.5

15        (92,77)            (92+77)/2=84.5

c.

The sample means of sample mean μxbar will calculated by taking average of sample means

μxbar=(71.5+ 81.5+ 80.5+ 85.5+ 78+ 74+ 73+ 78+ 70.5+ 83+ 88+ 80.5+ 87+ 79.5+ 84.5)/15

μxbar=1195/15=79.67

Population mean=μ=(79+64+84+82+92+77)/6

μ=478/6=79.67

Sample means of sample mean μxbar=Population mean μ.

5 0
3 years ago
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