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navik [9.2K]
3 years ago
9

One school survey showed that 3 out of 5 students own a pet. Another survey showed that 6 out of 11 students owned a pet. Are th

ese results equivalent? Explain your reasoning.
Mathematics
2 answers:
leonid [27]3 years ago
4 0
3/5 own a pet means that (0.6) or 60% own a pet
6/11 own a pet means that (0.545) or 54.5% own a pet

Conclusion, in the 1st class more people 60% own a pet whereas in the second class only 54.5% have a pet

Oxana [17]3 years ago
3 0
No they are not because 3/5 is not equal to 6/11. if they were equal, it would be 6/10
You might be interested in
−2x=x^2−6
Iteru [2.4K]

Step-by-step explanation:

Example 1

Solve the equation x3 − 3x2 – 2x + 4 = 0

We put the numbers that are factors of 4 into the equation to see if any of them are correct.

f(1) = 13 − 3×12 – 2×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2

f(2) = 23 − 3×22 – 2×2 + 4 = −4

f(−2) = (−2)3 − 3×(−2)2 – 2×(−2) + 4 = −12

f(4) = 43 − 3×42 – 2×4 + 4 = 12

f(−4) = (−4)3 − 3×(−4)2 – 2×(−4) + 4 = −100

The only integer solution is x = 1. When we have found one solution we don’t really need to test any other numbers because we can now solve the equation by dividing by (x − 1) and trying to solve the quadratic we get from the division.

Now we can factorise our expression as follows:

x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0

It now remains for us to solve the quadratic equation.

x2 − 2x − 4 = 0

We use the formula for quadratics with a = 1, b = −2 and c = −4.

We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:

x = 1

x = 1 + Ö5

x = 1 − Ö5

Example 2

We can easily use the same method to solve a fourth degree equation or equations of a still higher degree. Solve the equation f(x) = x4 − x3 − 5x2 + 3x + 2 = 0.

First we find the integer factors of the constant term, 2. The integer factors of 2 are ±1 and ±2.

f(1) = 14 − 13 − 5×12 + 3×1 + 2 = 0 1 is a solution

f(−1) = (−1)4 − (−1)3 − 5×(−1)2 + 3×(−1) + 2 = −4

f(2) = 24 − 23 − 5×22 + 3×2 + 2 = −4

f(−2) = (−2)4 − (−2)3 − 5×(−2)2 + 3×(−2) + 2 = 0 we have found a second solution.

The two solutions we have found 1 and −2 mean that we can divide by x − 1 and x + 2 and there will be no remainder. We’ll do this in two steps.

First divide by x + 2

Now divide the resulting cubic factor by x − 1.

We have now factorised

f(x) = x4 − x3 − 5x2 + 3x + 2 into

f(x) = (x + 2)(x − 1)(x2 − 2x − 1) and it only remains to solve the quadratic equation

x2 − 2x − 1 = 0. We use the formula with a = 1, b = −2 and c = −1.

Now we have found a total of four solutions. They are:

x = 1

x = −2

x = 1 +

x = 1 −

Sometimes we can solve a third degree equation by bracketing the terms two by two and finding a factor that they have in common.

6 0
3 years ago
Read 2 more answers
12) Given f = {(-3,40),(-2,25),(-1,14),(0,7),(1,4),(2,5),(3,7)}
never [62]

Answer:

<em>Part a) </em>Domain of f : {-3, -2, -1, 0, 1, 2, 3}

<em>Part b) </em>Domain of g : {-1, 0, 1, 2, 3, 4}

<em>Part c)  </em>Domain of f+g = {-1, 0, 1, 2, 3}

<em>Part d) </em>Ordered Pairs of f-g = {(-1, 10), (0, 2), (1, -2), (2, 4), (3, 23)}

Step-by-step explanation:

<em>Part a) Determining the domain of f </em>

Given f = {(-3,40),(-2,25),(-1,14),(0,7),(1,4),(2,5),(3,7)}

Domain is the set of the input values of x which define the function. In other words, domain is the set of all the first elements of order pairs.

Domain of f : {-3, -2, -1, 0, 1, 2, 3}

<em>Part b) Determining the domain of g</em>

Given g= {(-1,4),(0,5),(1,6),(2,1),(3,-16),(4,-51)}

As domain is the set of the input values of x which define the function. In other words, domain is the set of all the first elements of order pairs.

Domain of g : {-1, 0, 1, 2, 3, 4}

<em>Part c) Determining the domain of f+g</em>

<em>When there is a sum of two functions f and g, then domain of f+g will be the intersection of their domains.</em>

<em>As,</em>

<em>      </em>Given f = {(-3,40),(-2,25),(-1,14),(0,7),(1,4),(2,5),(3,7)}

      Domain of f : {-3, -2, -1, 0, 1, 2, 3}

and,

      Given g= {(-1,4),(0,5),(1,6),(2,1),(3,-16),(4,-51)}

       Domain of g : {-1, 0, 1, 2, 3, 4}

<em>As</em> when <em>there is a sum of two functions f and g, then domain of f+g will be the intersection of their domains</em>

So, the domain of f+g = {-1, 0, 1, 2, 3}

<em>Part d) List the ordered pairs of f-g</em>

As

    f = {(-3,40),(-2,25),(-1,14),(0,7),(1,4),(2,5),(3,7)}

and

    g = {(-1,4),(0,5),(1,6),(2,1),(3,-16),(4,-51)}

For f - g, we must focus on subtracting the second (y) coordinates of both function that correspond to the same element in the domain (x)

(f - g)(x) = f(x) - g(x)

(f - g)(x) = f(-1) - g(-1)  = 14 - 4 = 10

(f - g)(x) = f(0) - g(0)  = 7 - 5 = 2

(f - g)(x) = f(1) - g(1)  = 4 - 6 = -2

(f - g)(x) = f(2) - g(2)  = 5 - 1 = 4

(f - g)(x) = f(3) - g(3)  = 7 - (-16) = 23

So,

Ordered Pairs of f-g = {(-1, 10), (0, 2), (1, -2), (2, 4), (3, 23)}

Keywords:  domain, function, f+g, f-g

Learn more about domain, and ordered pairs from brainly.com/question/11422136

#learnwithBrainly

7 0
3 years ago
Simultaneously x-2y=0 40-y=14​
harkovskaia [24]

Answer:

x = 52, y = 26

Step-by-step explanation:

x - 2y = 0

40 - y = 14​

(40 - 40) - y = 14​ - 40

- y = -26

y = 26

x - 2(26) = 0

x - 52 = 0

x (- 52 + 52) = 0 + 52

x = 52

7 0
3 years ago
If your algorithm runs its critical section 1 + 2 + 3 + ... + (n-2) + (n-1) + n times, what is the asymptotic behavior of the al
Paraphin [41]
If 1 and 2 becomes multiplayed gave 3 and 3+3=4 (4-2)+(4-1)= 2+3=4 and 4 is n.
4 0
3 years ago
You draw 4 cards from a deck of 52 cards with replacement. What are the probabilities of drawing a black card on each of your fo
lawyer [7]

Solution:

As, in a pack of cards, there are 26 black cards and 26 red cards.

Probability of an event = \frac{\text{total favorable outcome}}{\text{total possible outcome}}

Probability of drawing a black card from a pack of 52 cards = \frac{26}{52}=\frac{1}{2}

As each card drawn is replaced,Each black card draw is an independent with another black card draw.

Probabilities of drawing a black card on each of  four trial=

  \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=[\frac{1}{2}]^4

4 0
3 years ago
Read 2 more answers
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