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NNADVOKAT [17]
3 years ago
9

What number us 17 less than the product of 7 and 11

Mathematics
1 answer:
m_a_m_a [10]3 years ago
3 0

77 is the product of 7 and 11 (Multiply them)

77 - 17 = 60

Hope this helped!

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The following data gives the scores of 13 students in a Biostatistics exam. 75 80 28 70 95 82 75 64 61 90 81 65 91 a) Find the f
Radda [10]

Answer:The following data gives the scores of 13 students in a Biostatistics exam. 75 80 28 70 95 82 75 64 61 90 81 65 91 a) Find the following statistical measures 1. Mean 2. Median 3. Mode 4. Range 5. 34th percentile 6. Interquartile Range 7. Variance 8. Standard deviation PRINCIPLES OF STATISTICS Assignment (1) Due Date: 15/7/2020 9. Coefficient of variation. b) (Without Calculations) If the instructor decide to add up 5 marks for every student, what are the values of the statistical measures mentioned in part (a). c) Construct the Boxplot of students' scores, and identify any possible outliers.

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3 years ago
Suppose you are putting up a tent for a camping trip.
SSSSS [86.1K]

The length of the rope needed is 15 feet

Given the following parameters;

  • Height of the pole = 9 foot
  • The distance from the <u>rope away from the base </u>of the pole is 12 feet

Required

  • Length of the rope.

To get the length of the rope, you will use the pythagoras theorem expressed as:

  • c² = a² + b²

Substituting the given parameters:

c² = 9² + 12²

c² =81 + 144

c² = 225

c = √225

c = 15 feet

Therefore the length of the rope needed is 15 feet

Learn more on Pythagoras theorem here: brainly.com/question/231802

7 0
2 years ago
Problem solving: Jade ran 6 times around her neighborhood to complete a total of 1 mile. How many times will she need to run to
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5 times, your welcome my friend
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3 years ago
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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.

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3 years ago
Relation sheep between the value of 6s in 6600
forsale [732]

The first 6 is equal to 6,000. The second 6 is equal to 600. 6,000 is 10 times greater than 600.

Hope this Helps!

3 0
3 years ago
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