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2 years ago

How many solutions can be found for the set of linear equations? Y = 7(x + 2) and y = 7x + 14

Mathematics

2 answers:

bixtya [17]2 years ago

5 0

Answer:

It is Infinite Solution

Step-by-step explanation:

NOOOO

I just did it on my Test and got a hundred

Bogdan [553]2 years ago

4 0

Answer:

infinite solution

Step-by-step explanation:

because Y = 7(x + 2) and y = 7x + 14 are the same

y = 7(x+2) = 7x + 14

eliminate to see

y = 7x + 14

-(y = 7x +14)

= 0 = 0 + 0

so it is infinite solution

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The given table values represents a liner equation. What is the slope?

Arada [10]

Answer:

A. 2

Step-by-step explanation:

$slope (m) = \frac{y_2 - y_1}{x_2 - x_1}$

Use any two points given in the table of values. Let's use (1, 5), (3, 9)

Let,

$(1, 5) = (x_1, y_1)$

$(3, 9) = (x_2, y_2)$

Plug these values into the slope formula:

$slope (m) = \frac{9 - 5}{3 - 1}$

$slope (m) = \frac{4}{2}$

$slope (m) = 2$

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3 years ago

What are the perfect numbers between 1 and 28?

mestny [16]

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7 0

2 years ago

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

2 years ago

6+9=?<img src="https://tex.z-dn.net/?f=6x%5E%7B2%7D%20%2B7x-9%3D" id="TexFormula1" title="6x^{2} +7x-9=" alt="6x^{2} +7x-9=" ali

ELEN [110]

Answer:

are u sure thet write the question right?

4 0

2 years ago

In hopes of encouraging healthier snacks at school, Bert brought in a tray of carrot sticks and apple slices to share. He brough

Oksi-84 [34.3K]

Answer:

40 % of the snacks were carrot sticks.

Step-by-step explanation:

Total number of carrot sticks in lunch = 18

Total number of apple slices in lunch = 27

So, Total Snacks in the Lunch = Carrot Sticks + Apple Slices

= 18 + 27 = 45

Now,

⇒Percentage of Carrot Sticks = 40%

Hence, 40 % of the snacks were carrot sticks.

4 0

2 years ago