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elena55 [62]
3 years ago
15

I need help with these three questions someone help me please what are these shapes called

Mathematics
1 answer:
lora16 [44]3 years ago
7 0

Answer:

The first shape is called (I don't know)

The second shape is called Icosahedron

The thrid shape is called Octahedron

Step-by-step explanation:

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The correct option is B- second

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Systems of equations, solve by substitution please!<br> x + 7y = -37<br> 4x - 3y = 7
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Answer:

1. x = -37 - 7y

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Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:
Vikentia [17]

Answer:

a

   \= x  = 18.5  ,  \sigma =  5.15

b

 15.505 < \mu <  21.495

c

 14.93 < \mu <  22.069

Step-by-step explanation:

From the question we are are told that

    The  sample data is  21, 14, 13, 24, 17, 22, 25, 12

     The sample size is  n  = 8

Generally the ample mean is evaluated as

        \= x  =  \frac{\sum x  }{n}

        \= x  =  \frac{  21 + 14 + 13 + 24 + 17 + 22+ 25 + 12  }{8}

         \= x  = 18.5

Generally the standard deviation is mathematically evaluated as

         \sigma =  \sqrt{\frac{\sum (x- \=x )^2}{n}}

\sigma =  \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}

\sigma =  5.15

considering part b

Given that the confidence level is  90% then the significance level is evaluated as

         \alpha  =  100-90

         \alpha  = 10\%

         \alpha  = 0.10

Next we obtain the critical value of  \frac{ \alpha }{2}  from the normal distribution table the value is  

     Z_{\frac{ \alpha }{2} }  =  1.645

The margin of error is mathematically represented as

      E =  Z_{\frac{ \alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>    E =1.645  *  \frac{5.15 }{\sqrt{8} }

=>     E =  2.995

The 90% confidence interval is evaluated as

       \= x  -  E < \mu <  \= x +  E

substituting values

       18.5 -  2.995 < \mu <  18.5 +  2.995

       15.505 < \mu <  21.495

considering part c

Given that the confidence level is  95% then the significance level is evaluated as

         \alpha  =  100-95

         \alpha  = 5\%

         \alpha  = 0.05

Next we obtain the critical value of  \frac{ \alpha }{2}  from the normal distribution table the value is  

     Z_{\frac{ \alpha }{2} }  =  1.96

The margin of error is mathematically represented as

      E =  Z_{\frac{ \alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>    E =1.96  *  \frac{5.15 }{\sqrt{8} }

=>     E = 3.569

The 95% confidence interval is evaluated as

       \= x  -  E < \mu <  \= x +  E

substituting values

       18.5 - 3.569 < \mu <  18.5 +  3.569

       14.93 < \mu <  22.069

8 0
3 years ago
Suppose that the probability of a baseball player getting a hit in an at-bat is 0.3089. If the player has 25 at-bats during a we
klio [65]

Answer:

P(X>9) = 0.3593

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=25, p=0.3089)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

For this case we want this probability:

P(X >9)

And we can use the complement rule like this:

P(X>9) = 1-P(X \leq 8)= 1-[P(X=0) + P(X=1) +....+P(X=8)]And we can find the individual probabilities like this:

P(X=0) =(25C0)(0.3089)^0 (1-0.3089)^{25-0} =0.0000974  

P(X=1) =(25C1)(0.3089)^1 (1-0.3089)^{25-1} =0.0011  

P(X=2) =(25C2)(0.3089)^2 (1-0.3089)^{25-2}=0.00584  

P(X=3) =(25C3)(0.3089)^3 (1-0.3089)^{25-3}= 0.02  

P(X=4) =(25C4)(0.3089)^4 (1-0.3089)^{25-4}=0.049  

P(X=5) =(25C5)(0.3089)^5 (1-0.3089)^{25-5}=0.092  

P(X=6)=(25C6) (0.3089)^6 (1-0.3089)^{25-6} = 0.138

P(X=7) =(25C7)(0.3089)^7 (1-0.3089)^{25-7}=0.167

P(X=8) =(25C8)(0.3089)^8 (1-0.3089)^{25-8}=0.168

And in order to do the operations we can use the following excel code:

"=1-BINOM.DIST(8,25,0.3089,TRUE)"  

And we got:

P(X>9) = 0.3593

4 0
3 years ago
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