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3 years ago

Can Someone Help Me Simplify These Expressions

Mathematics

1 answer:

Citrus2011 [14]3 years ago

7 0

-2k - (-7)k
= -2k + 7k
= 5k

-7(p + 2) + 9(p - 4)
= -7p - 14 + 9p - 36
= 2p - 50

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P is the probability of a certain event. Which of the following indicates that P is certain to happen?

podryga [215]

Note for probability, if it will never happen the P = 0. If it must happen P = 1.

So certain to happen:

C. P=1

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3 years ago

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What is the probability that the two pointers will land on an odd number and letter C? Give the probability as a percent

Firlakuza [10]

Answer:

12.5%

Step-by-step explanation:

hope this helps and its right

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The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

3 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

2 years ago

OK here's the deal, I need to graph F(x)=x^2 and F(x)=2^x as functions. I've run completely out of options, I've graphed them as

butalik [34]

Answer:

see attached graph

Step-by-step explanation:

You graphs look fine. The only thing I can think of without seeing the question itself, its that perhaps the teacher wants them on the same graph?

7 0

3 years ago