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JulijaS [17]
2 years ago
13

A rectangular football field is 64 meters wide and 100 meters long. A player runs from one corner of the field in a diagonal lin

e to the opposite corner. How far did the player run? Round your answer to the nearest meter.
Mathematics
2 answers:
joja [24]2 years ago
8 0
Let the distance be d

Using pythagorean theorem,

d^2 = 64^2 + 100^2
d^2 = 14096
d = 118.7
d ≈ 119 


The player ran for 119 meters
galben [10]2 years ago
8 0

Answer:

119

Step-by-step explanation:

its correct on khan

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5. When looking at a map, a student realizes that Birmingham is nearly due west of Atlanta, and Nashville is nearly due north of
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Answer: 250 mi

Step-by-step explanation:

Here we can think in a triangle rectangle:

The distance from Birmingham to Atlanta is roughly 150 mi, and this is one of the cathetus.

And the distance from Birmingham to Nashville is roughly 200 mi, this is the other cathetus of the triangle.

Now, the distance from Atlanta to Nashville will be the hypotenuse of this triangle rectangle.

Now we can apply the Pythagorean's theorem:

A^2 + B^2 = H^2

Where A and B are the cathetus, and H is the hypotenuse:

Then:

H = √(A^2 + B^2)

H = √(150^2 + 200^2) mi = √(62,500) mi = 250 mi

Then the estimated distance from Atlanta to Nashville is 250 mi

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2 years ago
Read the numbers and decide what the next number should be. 256 196 144 100 64 36 16
Tatiana [17]
340? TBD add the two numbers.
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4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln
pochemuha

Answer:

No these these result do not differ at 95% confidence level  

Step-by-step explanation:

From the question we are told that

  The first concentrations is  c _1= 30.0 \ g/m^3

      The second concentrations is  c _2 = 52.9 \ g/m^3

  The first sample size is  n_1 =  32

    The second sample size is  n_2 =  32

   The  first standard deviation is \sigma_1 =  30.0

     The  first standard deviation is \sigma_1 =  29.0

The mean for  Turnpike is  \= x _1 = \frac{c_1}{n}  =  \frac{31.4}{32} = 0.98125

The mean for   Tunnel is  \= x _2 = \frac{c_2}{n}  =  \frac{52.9}{32} = 1.6531

The  null hypothesis is  H_o  :  \mu _1 - \mu_2  =  0

The  alternative hypothesis is  H_a  :  \mu _1 - \mu_2  \ne  0

Generally the test statistics is mathematically represented as

              t =  \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1}  +\frac{\sigma_2^2}{n_2} }}

         t =  \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32}  +\frac{29^2}{32} }}

        t = - 0.0899

Generally the degree of freedom is mathematically represented as

     df =  32+ 32 - 2

     df =  62

The  significance \alpha  is  evaluated as

      \alpha  =  (C - 100 )\%

=>   \alpha  =  (95 - 100 )\%

=>   \alpha  =0.05

The  critical value  is evaluated as

      t_c  =  2  *  t_{0.05 ,  62}

From the student t- distribution table  

        t_{0.05, 62} =  1.67

So

     t_c  =  2 * 1.67

=>  t_c  = 3.34

given that

       t_c  >  t we fail to reject the null hypothesis so  this mean that the result do  not differ

       

6 0
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Answer:

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