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3 years ago

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John thinks that any two lines must have a point of intersection. Is he correct? If so how do you know. If not, produce a counte

rexample.

2 answers:

katrin [286]3 years ago

8 0

Answer: parallel lines

bekas [8.4K]3 years ago

6 0

This isn't necessarily true. When two lines have equivalent slopes, then they are parallel and will never intersect.

y=1/2x + 12
y=1/2x + -2
These lines can't intersect because there are equal slopes.

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Help me please help me help please me please pleases me please please help

kifflom [539]

Answer:

the answer is 15/14 when you add them

7 0

3 years ago

Read 2 more answers

Find -48a divided by 8 when a=1 write your answer in simplest form

lukranit [14]

-48a÷ 8 can be written as,

$\frac{-48a}{8}$

Put a=1 in the above exprression and solve.

$\frac{-48\times1}{8}=-6$

Therefore, the solution is -6.

8 0

1 year ago

Marty made a $220 bank deposit using $10 bills and $5 bills. She gave the teller a total of 38 bills, how many $5 bills were in

Luda [366]

ANSWER: 32 five-dollar bills

======

EXPLANATION:

Let x be number of $5 bills

Let y be number of $10 bills

Since we have total of 38 bills, we must have the sum of x and y be 38

x + y = 38 (I)

Since the total amount deposited is $220, we must have the sum of 5x and 10y be 220 (x and y are just the "number of" their respective bills, so we multiply them by their value to get the total value):

5x + 10y = 220 (II)

System of equations:

$\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ 5x + 10y &= 220 && \text{(II)} \end{aligned} \right.$

Divide both sides of equation (II) by 5 so our numbers become smaller

$\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ x + 2y &= 44 && \text{(II)} \end{aligned} \right.$

Rearrange (I) to solve for y so that we can substitute into (II)

$\begin{aligned} x + y &= 38 && \text{(I)} \\ y &= 38 - x \end{aligned}$

Substituting this into equation (II) for the y:

$\begin{aligned} x + 2y &= 44 && \text{(II)} \\ x + 2(38 - x) &= 44\\ x + 76 - 2x &= 44 \\ -x &= -32 \\ x &= 32 \end{aligned}$

We have 32 five-dollar bills

======

If we want to finish off the question, use y = 38 - x to figure out number of $10 bills

$y = 38 - 32 = 6$

32 five-dollar bills and 6 ten-dollar bills

7 0

3 years ago

Consider the graph of the function.

Zina [86]

Answer:B

looks like it has asymptotes but doesnt actually

6 0

2 years ago

It is estimated that 75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell

Mademuasel [1]

Answer:

a) 75

b) 4.33

c) 0.75

d) $3.2 \times 10^{-13}$ probability that no one in a simple random sample of 100 young adults owns a landline

e) $6.2 \times 10^{-61}$ probability that everyone in a simple random sample of 100 young adults owns a landline.

f) Binomial, with $n = 100, p = 0.75$

g) $4.5 \times 10^{-8}$ probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they do not own a landline, or they do. The probability of an young adult not having a landline is independent of any other adult, which means that the binomial probability distribution is used 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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.

This means that $p = 0.75$

(a) On average, how many young adults do not own a landline in a random sample of 100?

Sample of 100, so $n = 100$

$E(X) = np = 100(0.75) = 75$

(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100(0.75)(0.25)} = 4.33$

(c) What is the proportion of young adults who do not own a landline?

The estimation, of 75% = 0.75.

(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?

This is P(X = 100), that is, all do not own. So

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

$P(X = 100) = C_{100,100}.(0.75)^{100}.(0.25)^{0} = 3.2 \times 10^{-13}$

$3.2 \times 10^{-13}$ probability that no one in a simple random sample of 100 young adults owns a landline.

(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?

This is P(X = 0), that is, all own. So

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

$P(X = 0) = C_{100,0}.(0.75)^{0}.(0.25)^{100} = 6.2 \times 10^{-61}$

$6.2 \times 10^{-61}$ probability that everyone in a simple random sample of 100 young adults owns a landline.

(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?

Binomial, with $n = 100, p = 0.75$

(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?

This is P(X = 50). So

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

$P(X = 50) = C_{100,50}.(0.75)^{50}.(0.25)^{50} = 4.5 \times 10^{-8}$

$4.5 \times 10^{-8}$ probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

8 0

3 years ago