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aleksklad [387]
4 years ago
10

Set up an equation that can be used to solve the problem. Solve the equation and answer the question asked.

Mathematics
1 answer:
BlackZzzverrR [31]4 years ago
3 0

I believe the answer is D

Showing my work i did 224m+196m= 420 Miles

420 divided by 7 equals 60mph


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A pool measuring 16 meters by 22 meters is surrounded by a path of uniform​ width, as shown in the figure. If the area of the po
lesantik [10]

Let X be the width of the path.

The width of the path would be the width of the pool plus 2x, so 2x +16

The length of the path would be the length of the pool plus 2x, so 2x+22

The total area is 832 square feet.

Area is found by multiplying the length by the width.

So you have:

(2x+16) (2x+22) = 832

Use the Foil Method:

(2x+16) (2x+22) = 4x^2 +76x+352

Now you have:

4x^2 +76x+352 = 832

Subtract 832 from both sides:

4x^2 +76x+352-832 = 0

Simplify:

4x^2 +76x -480

Factor the left side:

4(x-5)(x+24) = 0

Divide both sides by 4:

(x-5)(x+24) = 0

Set both parenthesis to equal 0 and solve for x:


x=5 and x = -24

The width of the path cannot be a negative number, so the width is 5 meters.


8 0
3 years ago
Help !!!!!!!!!!!!!!!
Ostrovityanka [42]

Answer:

  x = 29

Step-by-step explanation:

Follow the directions: substitute 13 for y in the first equation.

  x + 13 = 42

Subtract 13 from both sides of the equation.

  x +13 -13 = 42 -13

  x = 29

4 0
3 years ago
Given m||n, find the value of x.
kkurt [141]

As angle x° and angle 76° are alternate interior angles their measure will be equal.

Which means :-

x°  =  76°  \:

Therefore , the value of x = 76° .

7 0
3 years ago
A monarchy has one leader how many terms do you think a monomial has
Sholpan [36]
I would think i has one term because the prefix "mono" means one
3 0
3 years ago
Read 2 more answers
Bodytemperatures ofhealthy koalasarenormally distributed with a mean of 35.6°C and a standard deviation of 1.3°C. a.What is the
Licemer1 [7]

Answer:

a

 P(X  < 35) =  32.3 \%

b

P(\= X  < 35) =  0.6 \%

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

c

A potential confounding variable for this study is  the population of the koalas because in the first question the population was not taken into account and the probability was  P(X  < 35) =  32.3 \% but when the population was taken into account (i.e  n =  30) the probability became

P(\= X  < 35) =  0.6 \%  

Step-by-step explanation:

From the question we are told that

  The mean is  \mu =  35.6^oC

   The standard deviation is  s =  1.3^oC

   The sample size is  n = 30

Generally the  probability that a health koala has a body temperature less than 35.0°C is mathematically represented as

     P(X  < 35) =  P(\frac{X  - \mu }{s}  < \frac{35 - 35.6}{1.3} )

Here  (\frac{X  - \mu }{s} =  Z (The   \ standardized \  value \  of  \  X )

So

    P(X  < 35) =  P(Z < -0.46)

From the z-table  P(Z <  -0.46) =  0.323

So  

    P(X  < 35) =  0.323

Converting to percentage

      P(X  < 35) =  0.323  * 100

      P(X  < 35) =  32.3 \%

considering question b

The sample mean is  \= x =  35

Generally the standard error of the mean is mathematically represented as

   \sigma_{\= x} =  \frac{s}{\sqrt{n} }

=>  \sigma_{\= x} =  \frac{1.3}{\sqrt{30} }

=>  \sigma_{\= x} =  0.2373

Generally the probability of the mean body temperature of koalas being less than 35.0°C is mathematically represented as

 P(\= X  < 35) =  P(\frac{\= X  -  \mu  }{\sigma_{\= x }} <  \frac{35 -35.6}{0.2373 }  )

P(\= X  < 35) =  P(Z< -2.53  )

From the z-table  we have that

   P(Z< -2.53  ) =  0.006

So

 P(\= X  < 35) = 0.006 /tex] Converting to percentage        [tex]P(\= X  < 35) =   0.006  * 100

      P(\= X  < 35) =  0.6 \%

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

A potential confounding variable for this study is  the population of the koalas because in the first question the population was not taken into account and the probability was  P(X  < 35) =  32.3 \% but when the population was taken into account (i.e  n =  30) the probability became

 P(\= X  < 35) =  0.6 \%  

4 0
4 years ago
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