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Olin [163]
3 years ago
7

X + 16 = 9z Help please

Mathematics
2 answers:
kompoz [17]3 years ago
7 0

Answer:

Hallooooooo

Step-by-step explanation:

diamong [38]3 years ago
3 0

Answer:x=9z + 16

                  z= x/9 + 16/9

Step-by-step explanation: dont know if i solve for x or z so i did both  subtract 16 from both sides for x      

Isolate the variable by dividing each side by factors that don't contain the variable. for z let me know if u need help

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Show that the set of functions from the positive integers to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is uncountable. [Hint: First
Sedbober [7]

Answer:

since the set of functions expressed are uncountable and they are a subset of real numbers starting from N therefore the set {0,1,2,3,4,5,6,7,8,9} is uncountable as well as its off functions

Step-by-step explanation:

set = {0,1,2,3,4,5,6,7,8,9}

setting up a one-to-one correspondence between the set of real numbers between 0 and 1

The function : F(n)= {0,1} is equivalent to the subset (sf) of (n) , this condition is met if n belongs to the subset (sf) when f(n) = 1

hence The power set of (n) is uncountable and is equivalent to the set of real numbers given

since the set of functions expressed are uncountable and they are a subset of real numbers starting from N therefore the set {0,1,2,3,4,5,6,7,8,9} is uncountable as well as its offfunctions

3 0
3 years ago
Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.
Sindrei [870]

Answer:

a) P(X=3) = 0.1

b) P(X\geq 3) =1-P(X

And replacing we got:

P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4

c) P(X=4) = 0.3

d) P(X=0) = 0.2

e) E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2

f) E(X^2)= \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4

And the variance would be:

Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4

And the deviation:

\sigma =\sqrt{2.4} = 1.549

Step-by-step explanation:

We have the following distribution

x      0     1     2   3   4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

P(X=3) = 0.1

Part b

We want this probability:

P(X\geq 3) =1-P(X

And replacing we got:

P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4

Part c

For this case we want this probability:

P(X=4) = 0.3

Part d

P(X=0) = 0.2

Part e

We can find the mean with this formula:

E(X)= \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2

Part f

We can find the second moment with this formula

E(X^2)= \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4

And the variance would be:

Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4

And the deviation:

\sigma =\sqrt{2.4} = 1.549

4 0
4 years ago
What's the answer ????
Citrus2011 [14]
The third one (I think) because she gets 20$ every week
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Please help on these questions and explain please!!
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The one in the middle is 12 for the median = 8
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3 years ago
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PLZ ASAP, I NEED HELP ASAP
loris [4]
I won't give you the answer because I am sure you are capable of finding it yourself once given a push, but what I would do is simplify all of the numbers into a mixed fraction (a b/c) and then go from there. Don't subtract the two numbers with the denominator of 25 first because they goes against PEMDAS. Change all of the fractions so they have like denominators, and once you solve it, just write the fraction part of the answer (b/c). If you have any other questions, just ask.
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