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frutty [35]
3 years ago
8

Help me on number two plz

Mathematics
1 answer:
s2008m [1.1K]3 years ago
7 0

Okay so Sam has completed 5/13 of the crossword puzzle and wants to do 8/13.

So we need to find out how many more she wants to do.

8/13-5/13=3/13

She has 3/13 has to complete.

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In a certain​ country, the mean birth weight for boys is 3.27 ​kg, with a standard deviation of 0.51 kg. Assuming that the distr
Degger [83]

Answer:

0.066

Step-by-step explanation:

We solve this question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation

For the question

mean birth weight for boys is 3.27 ​kg, with a standard deviation of 0.51 kg.

x = 2.5

Hence:

z = 2.5 - 3.27/0.51

z = -1.5098

Probability value from Z-Table:

P(x ≤ 2.5) =P(x < 2.5) = P(x = 2.5) =

=0.065547

Therefore, the proportion of baby boys that are born with a low birth weight is 0.066

5 0
3 years ago
Parallelogram abcd is shown below.
Ilya [14]

the answer is C.DAB and BCD

4 0
3 years ago
For x, y ∈ R we write x ∼ y if x − y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1)
vodomira [7]

Answer:

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

  • The relation being reflexive (∀a∈R, a∼a)
  • The relation being symmetric (∀a,b∈R, a∼b⇒b∼a)
  • The relation being transitive (∀a,b,c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it´s known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, -k ∈ Z

b∼a, ∀a,b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a,b,c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k,l ∈ Z

(a-b)+(b-c)=k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a,b,c ∈ R

∼ is Transitive by definition

We´ve shown that ∼ is an equivalence relation on R.

B. Now we have to show that there´s a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F(x)=[x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x,y∈[0,1), F(x)=F(y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F(x)=b):

F is injective:

let x,y ∈ [0,1) and suppose F(x)=F(y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x,y ∈ [0,1), then k must be 0. If it isn´t, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x,y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let´s find x such as x ∈ [0,1) and F(x)=[b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let´s show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F(r)=[b]

∼ is surjective

Then F maps [0,1) into C, i.e [0,1) is a set of representatives for the set of the equivalence classes.

4 0
2 years ago
Which expression represents the volume of the prism?
Tamiku [17]

Answer:

x³-2x²  = x2(x – 2) cubic units

Step-by-step explanation:

The volume of a prism is found by multiplying the base area by the height. The base area is a parallelogram and so the area is x*(x-2) = x² -2x.

Multiply this area by the height x.

V = x(x² - 2x) = x³-2x²

This is the same as x2(x – 2) cubic units.

7 0
2 years ago
Read 2 more answers
How to get to the answer
Ilia_Sergeevich [38]

Answer:

\large\boxed{C)\ F(x)=2^x+1}

Step-by-step explanation:

C)\ F(x)=2^x+1\\\\\text{Put the values of x from the table to the equation of a function:}\\\\F(3)=2^3+1=8+1=9\qquad\ CORREC\\F(4)=2^4+1=16+1=17\qquad CORRECT\\F(6)=2^6+1=64+1=65\qquad CORRECT\\F(7)=2^7+1=128+1=129\qquad CORRECT\\F(8)=2^8+1=256+1=257\qquad CORRECT

7 0
2 years ago
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