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crimeas [40]
3 years ago
14

What is the range of the following list of ordered pairs? (-3, -1), (0, -2), (4, 3), (1, 5)

Mathematics
1 answer:
lawyer [7]3 years ago
6 0
(-3,-1),(0,-2),(1,5),(4,3)
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( 20 points + brainliest for the correct answer!!!) <br><br> ASA AAS CONGRUENCE
Olegator [25]
A. AAS
B. SAS
C. ASA

Just look at what order they come in. Lmk if you need further help.
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Marcus boxers dog 3/5 of a mile every day how far does Marcus walk his dog 6 days
lina2011 [118]
Marcus walks 3.5 miles
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Determine whether the normal sampling distribution can be used. The claim is p &lt; 0.015 and the sample size is n
Svetradugi [14.3K]

Complete Question

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n=150

Answer:

Normal sampling distribution can not be used

Step-by-step explanation:

From the question we are told that

    The  null hypothesis is  H_o  :  p =  0.015

     The  alternative hypothesis is    H_a  :  p  <  0.015

     

The  sample size is  n= 150

Generally in order to use  normal sampling distribution  

     The value  np  \ge  5

So  

         np =  0.015 * 150

         np =  2.25

Given that  np < 5   normal sampling distribution  can not be used

8 0
3 years ago
Write greatest unit fractions that are repeating decimals. Then express each fraction as a decimal.
LekaFEV [45]
Our repeating decimal would be= 0.5555….
Let’s give 0.5555….. repeating decimal a variable of x:
x= 0.5555…. or 0.5(move 1 point to the right)
So ,10x = 5.5555 or 5.5

Now, Let’s do subtraction:
10x = 5.5555… or 5.5  
<u>-   x =  0.5555… or 0.5</u>
 9x  = 5.0            or 5.0

To get a fraction, provide a denominator the same with the numerator of variable x which is 9. Then divide the difference:
<span><u>9x</u> = <u>5.0</u>  
9        9</span> <span> <span><span> <span> <span>Therefore, x = <u>5
</u>                                     9</span> </span> </span> </span></span>


7 0
3 years ago
(4.1.4) Let X and Y be Bernoulli random variables. Let Z = X + Y. a. Show that if X and Y cannot both be equal to 1, then Z is a
Fynjy0 [20]

Step-by-step explanation:

Given that,

a)

X ~ Bernoulli (p_x) and Y ~ Bernoulli (y_x)

X + Y = Z

The possible value for Z are Z = 0 when X = 0 and Y = 0

and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0

If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,

Therefore Z is a Bernoulli random variable

b)

If X and Y can not be both equal to  1

then,

p_z = P(X=1 or Y=1)\\

p_z = P(X=1)+P(Y=1)-P(=1 and Y =1)

p_z = P(x=1)+P(Y=1)\\\\p_z=p_x+p_y

c)

If both X = 1 and Y = 1 then Z = 2

The possible values of the random variable Z are 0, 1 and 2.

since a  Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution

7 0
3 years ago
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