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aleksandrvk [35]
3 years ago
15

The diagram shows a square of side 3 in. containing a circle of diameter 3 in. To the nearest hundredth, what is the area of the

shaded part of the figure? Use 3.14 for p.
Mathematics
1 answer:
Westkost [7]3 years ago
7 0
The area of the square is 9 square inches.
The area of the circle is about 7.065. As a result, the area of the shading region is 1.94 square inches. Hope it help!
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Please Help
Fofino [41]

Answer:

r

Step-by-step explanation:......

5 0
3 years ago
Round 314550 to 1 significant figure​
Sauron [17]
300000 all you had to do was round the number down after the 1 significant figure

you round down with
1,2,3,4

and round up with
5,6,7,8,9

so if it asked for 78453
to 2 significant figures you count the numbers and round after that
so 78|453 and since 4 is the next number and it is less that 5 we round down which means it becomes 78000 with is lower than 78453

but if we had 0.0247
and it asked for this number to 2 significant figures all you need to do is count the SIGNIFICANT figures significant means anything with value so that is 1,2,3,4,5,6,7,8,9,10.... 0 has no value therefore we ignore 0 and start counting at the first figure larger than 0
if we go back to 0.0247 that would be the 2 so 2 significant figure of 0.0247 would equal
0.024|7
7 is larger than 5 so we round up to
0.025
tadaa i hope that helped
8 0
3 years ago
Read 2 more answers
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

                   

3 0
3 years ago
Need help with my homework 2 problems
Solnce55 [7]
4(x+2)  ,   2(2x+4)  ,  8+4x  ,   1/2(8x+16) are all the same as 4x+ 8.
5 0
3 years ago
Can I get help with this pls
LuckyWell [14K]

Answer:

c

Step-by-step explanation:

7 0
3 years ago
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