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zepelin [54]
3 years ago
11

At the beginning of a​ walk, roberto and juanita are 7.77.7 miles apart. if they leave at the same time and walk in the same​ di

rection, roberto overtakes juanita in 11 hours. if they walk towards each​ other, they meet in 1 hour. what are their​ speeds
Mathematics
1 answer:
dem82 [27]3 years ago
4 0
Roberto overtakes Juanita at the rate of (7.7 mi)/(11 h) = 0.7 mi/h. This is the difference in their speeds. The sum of their speeds is (7.7 mi)/1 h) = 7.7 mi/h.

Roberto walks at the rate (7.7 + 0.7)/2 = 4.2 mi/h.
Juanita walks at the rate 4.2 - 0.7 = 3.5 mi/h.


_____
In a "sum and diference" problem, one solution is half the total of the sum and difference. If we let R and J be the respective speeds of Roberto and Juanita, we have
  R + J = total speed
  R - J = difference speed
Adding these two equations, we have
  2R = (total speed + difference speed)
  R = (total speed + difference speed)2 . . . . . as computed above
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The average of 3,15, and my number is 40. What is my number?
mylen [45]

Answer:

Your number is 102.

120/3=40 (3 numbers so divide by 3)

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8 0
3 years ago
1.Graph f(x) = -1.5x +6<br> 2.Graph f(x) = -1/2x-5
nalin [4]

Ans(1):

Given equation is f(x)=-1.5x+6

we can plug any number like x=0 and x=2 to find the f(x) also called y-value

plug x=0

f(x)=-1.5x+6 =-1.5*0+6 =0+6 =6

Hence first point is (0,6)

plug x=2

f(x)=-1.5x+6 =-1.5*2+6 =-3+6 =3

Hence first point is (2,3)

now we can graph both points then join them to get final graph of f(x)=-1.5x+6

---------------------

Ans(2):

We can repeat exactly same process for f(x) = -1/2x-5.

So the final graph will look like attached picture:


5 0
2 years ago
2564.0 millimeters tall. If the boards are too long they must be trimmed, and if the boards are too short they cannot be used. A
Leokris [45]

Answer:

We conclude that the board's length is equal to 2564.0 millimeters.

Step-by-step explanation:

We are given that a sample of 26 is made, and it is found that they have a mean of 2559.5 millimeters with a standard deviation of 15.0.

Let \mu = <u><em>population mean length of the board</em></u>.

So, Null Hypothesis, H_0 : \mu = 2564.0 millimeters    {means that the board's length is equal to 2564.0 millimeters}

Alternate Hypothesis, H_A : \mu \neq 2564.0 millimeters      {means that the boards are either too long or too short}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                             T.S.  =  \frac{\bar X-\mu}{\frac{s }{\sqrt{n}} }  ~  t_n_-_1

where, \bar X = sample mean length of boards = 2559.5 millimeters

            s = sample standard deviation = 15.0 millimeters

             n = sample of boards = 26

So, <em><u>the test statistics</u></em> =  \frac{2559.5-2564.0}{\frac{15.0 }{\sqrt{26}} }  ~   t_2_5

                                     =  -1.529    

The value of t-test statistics is -1.529.

Now, at a 0.05 level of significance, the t table gives a critical value of -2.06 and 2.06 at 25 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the board's length is equal to 2564.0 millimeters.

4 0
3 years ago
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