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kotykmax [81]
3 years ago
14

2(1 – x) > 2x HELP ASAP!!! I CANT FIGURE THIS OUT

Mathematics
1 answer:
katrin [286]3 years ago
4 0
<h3>Answer:  x < 1/2</h3>

This is the same as saying x < 0.5 since 1/2 = 0.5

====================================================

Work Shown:

2(1-x) > 2x

2*1 + 2*(-x) > 2x ..... distribute

2 - 2x > 2x

2 > 2x+2x .... adding 2x to both sides

2 > 4x

4x < 2 ..... flip both sides; flip the inequality sign

x < 2/4 ..... dividing both sides by 4; see note below

x < 1/2

Note: We do not flip the inequality sign here because we are dividing both sides by a positive value. The inequality sign would only flip if you divide both sides by a negative number.

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A basketball player is fouled while attempting to make a basket and receives two free throws. The opposing coach believes there
Elan Coil [88]
B not quiet sure but if anything try again : )
6 0
2 years ago
The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.
mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution  

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.96516 probability of a chipmunk living through the year, thus p = 0.96516

Item a:

  • Two is P(X = 2) when n = 2, thus:

P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

  • Six is P(X = 6) when n = 6, then:

P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

  • At least one not living is:

P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

6 0
2 years ago
A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting
Soloha48 [4]

Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= \frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees}  = \frac{6}{15}

b.) The probability that a secretary wins a ticket on second draw.  It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant  wins on first draw)

=\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}

= \frac{\frac{5}{15}  \times \frac{6}{14}}{\frac{5}{15} }  = \frac{6}{14}  .

The probability that  a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw  = \frac{6}{15} is not true.

The probability that a secretary wins on the second draw  = \frac{number \hspace{0.1cm} of  \hspace{0.1cm} secretaries  \hspace{0.1cm} remaining } { number  \hspace{0.1cm} of  \hspace{0.1cm} employees  \hspace{0.1cm} remaining}  = \frac{6 - 1}{15 - 1}  = \frac{5}{14}

c.) The probability that a consultant wins on the first draw  =

\frac{number \hspace{0.1cm} of  \hspace{0.1cm} consultants  \hspace{0.1cm}  } { number  \hspace{0.1cm} of  \hspace{0.1cm} employees  \hspace{0.1cm} }  = \frac{5 }{15}  = \frac{1}{3}

d.) The probability of two secretaries winning both tickets

= (probability of a secretary winning in the first draw) × (The probability that a secretary wins on the second draw)

= \frac{6}{15}  \times \frac{5}{14}  = \frac{1}{7}

Therefore the only statement that is not true is b.)

5 0
3 years ago
PLEASE PLEASE PLEASE HELP!!
irga5000 [103]
  9            a
------  =  -----
  a            5

a^2 = 45 

a = 3<span>√5
</span>
  4            b
------  =  -----
  b            5

b^2 = 20

b = 2√5

answer: 

a = 3√5
b = 2√5

hope it helps
7 0
2 years ago
PLZ PLZ PLZ HELP ME!!! AND QUICKLY TOO!!! (15 POINTS)
Orlov [11]

The answer would be:


A translation 5 units down, followed by a 180-degree counterclockwise rotation about the origin


See attached picture:

8 0
3 years ago
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