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taurus [48]
3 years ago
13

N + 25= 60 what is the value of n

Mathematics
2 answers:
gtnhenbr [62]3 years ago
7 0

Answer:

n=35

Step-by-step explanation:

Subtract 25 from both sides to isolate the variable

n+25=60

  -25  -25

n=35

Hope this helps! :)

Angelina_Jolie [31]3 years ago
6 0

Answer:

Step-by-step explanation:

N+25=60

N=60-25

N=35

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OlgaM077 [116]

Answer:

A- 2x + 8

Step-by-step explanation:

Combine like terms

6x - 4x = 2x. 2x + 8 is left

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3 years ago
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Suppose the expected tensile strength of type-A steel is 103 ksi and the standard deviation of tensile strength is 7 ksi. For ty
ExtremeBDS [4]

Answer:

a

i So  the approximate distribution of \= X is \mu_{\= X} =103 and  \sigma_{\= X} = 0.783

ii So the approximate distribution of \= Y is \mu_{\= Y} =105 and  \sigma_{\= Y} = 0.645

b

 the approximate distribution of  \=X  - \= Y is E (\= X - \= Y)  = -2 and  \sigma_{\= X  - \=Y}=1.029

Here we can see that the mean of the approximate distribution is negative which tell us that this negative value of the  data for  \=X  - \= Y sample   are more and their frequency occurrence is higher than the positive values  

c

the value of  P(-1 \le \=X - \= Y  \le 1) is = -0.1639    

Step-by-step explanation:

From the question we are given that

       The expected tensile strength of the type A steel is  \mu_A = 103 ksi

        The standard deviation of type A steel is  \sigma_A = 7ksi

         The expected tensile strength of the type B steel is \mu_B = 105\ ksi

            The standard deviation of type B steel is  \sigma_B = 5 \ ksi

Also the assumptions are

       Let \= X be the sample average tensile strength of a random sample of 80 type-A specimens

Here n_a =80

      Let \= Y be  the sample average tensile strength of a random sample of 60 type-B specimens.

  Here n_b = 60

Let the sampling distribution of the mean be

             \mu _ {\= X} = \mu

                   =103

 Let the sampling distribution of the standard deviation be

               \sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }

                     = \frac{7}{\sqrt{80} }

                    =0.783

So What this mean is that the approximate distribution of \= X is \mu_{\= X} =103 and  \sigma_{\= X} = 0.783

For \= Y

 The sampling distribution of the sample mean is

               \mu_{\= Y} = \mu

                    = 105

  The sampling distribution of the standard deviation is

               \sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }

                    = \frac{5}{\sqrt{60} }

                    = 0.645

So What this mean is that the approximate distribution of \= Y is \mu_{\= Y} =105 and  \sigma_{\= Y} = 0.645                      

Now to obtain the approximate distribution for \=X  - \= Y

               E (\= X - \= Y) = E (\= X) - E(\= Y)

                                =  \mu_{\= X} - \mu_{\= Y}

                                = 103 -105

                                = -2

The standard deviation of \=X  - \= Y is

               \sigma_{\= X  - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}

                         = \sqrt{(0.783)^2 + (0.645)^2}

                         =1.029

Now to find the value of  P(-1 \le \=X - \= Y  \le 1)

  Let us assume that F = \= X - \= Y

    P(-1 \le F \le 1) = P [\frac{-1 -E (F)}{\sigma_F} \le Z \le  \frac{1-E(F)}{\sigma_F} ]

                             = P[\frac{-1-(-2)}{1.029}  \le  Z \le  \frac{1-(-2)}{1.029} ]

                             =  P[0.972 \le Z \le 2.95]

                             = P(Z \le 0.972) - P(Z \le 2.95)

Using the z-table to obtain their z-score

                             = 0.8345 - 0.9984

                             = -0.1639

                   

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3 years ago
Suppose triangle TIP and triangle TOP are isosceles triangles. Also suppose that TI=5, PI=7, and PO=11. What are all the possibl
Ede4ka [16]
<h3>Two answers: 5, 7</h3>

====================================================

Explanation:

A drawing may be helpful to see what's going on. Check out the diagram below. This is one way of drawing out the two triangles. The locations of the points don't really matter, and neither does the the orientation of how you rotate things. What does matter is we have the right points connected to form the segments mentioned.

----------

For now, focus on triangle TIP only. In order to have this be isosceles, we must make TP = 5 or TP = 7.

If TP = 5, then it's the same length as TI.

If TP = 7, then it's the same length as PI.

In either case, we have exactly two sides the same length (the other side different) which is what it means for a triangle to be isosceles.

----------

Let's consider triangle TOP. For it to be isosceles, we must have two sides the same length. We already locked in TP to be either 5 or 7 in the previous section above. So there's no way that TP could be 11 units long to match up with PO = 11.

If TP = 5, then OT must also be 5 units long so that triangle TOP is isosceles.

If TP = 7, then OT = 7 for similar reasoning.

Either way, TP only has two choices on what it could be.

----------

In short, we basically just write the first two values given to us to get the two triangles to be isosceles. We can't use TP = 11 as it would make triangle TIP to be scalene (all sides are different lengths).

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