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ValentinkaMS [17]

3 years ago

5

Sandy cut a 151-centimeter board into five pieces. The three shortest pieces are the same length and the two longer pieces are e

ach three centimeters longer than a short piece. How long is each piece?

1 answer:

Serga [27]3 years ago

8 0

Answer: The short piece is 29.6 cm long and 32.6 cm long.

Step-by-step explanation:

Let x = length of short pieces , y = lengthof longer pieces.

y=x+3 (i)

As per given,

3x+2y=151 (ii)

Substitute value of y from (i) in (ii), we get

$3x+2(x+3)=151\\\\\Rightarrow\ 3x+2x+3=151\\\\\Rightarrow\ 5x=148\\\\\Rightarrow\ x=\dfrac{148}{5}=29.6$

Put value of in (i), we get

y=29.6+3= 32.6

Hence, the short piece is 29.6 cm long and 32.6 cm long.

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Simplify the expression. 8w + 5r – t + 9r

Wittaler [7]

Answer:

14r−t+8w

Step-by-step explanation:

Let's simplify step-by-step.

8w+5r−t+9r

=8w+5r+−t+9r

Combine Like Terms:

=8w+5r+−t+9r

=(5r+9r)+(−t)+(8w)

=14r+−t+8w

Answer:

=14r−t+8w

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A gate is made up of a rectangle and has a height of 6ft find the area of the gate

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You are at a European beach with 60 other visitors. 36 of them speak English. If you randomly meet two people on the beach, what

Kaylis [27]

Answer:

Assuming I'm one of the 36 English speakers and the other 24 speak Spanish for illustration purposes. The problem can be modeled as 59 marbles with 35 E and 24 S marbles as N = 59 possible outcomes = n(E) + n(S) = 35 + 24.

So I reach into the pile of marbles (on the beach) and the probability that it's p(E) = n(E)/N = 35/59 = 0.593220339 when I meet the one person. ANS

I assume I remember that first person; so I remove him from the marbles (by avoiding him on the beach) and now my probability is p(E and E) = n(E)/N * n(E)-1/(N - 1) = 35/59*34/58 = 0.347749854 ANS

Following the same logic p(E and E and E) = 35/59*34/58*33/57 = 0.201328863 ANS

This last one is different from the first three. This one is p(E >= 1|4 attempts). We can trace out a probability tree to identify those branches that contain at least one E event. So:

EEEE p() = 35/59 * 34/58 * 33/57 * 32/56 =

EEES p() = 35/59 * 34/58 * 33/57 * 24/56 =

EESE p() = 35/59 * 34/58 * 24/57 * 33/56 =

ESEE p() = 35/59 * 24/58 * 34/57 * 33/56 =

SEEE p() = 24/59 * 35/58 * 34/57 * 33/56 =

EESS p() = 35/59 * 34/58 * 24/57 * 23/56 =

ESES p() = 35/59 * 24/58 * 34/57 * 23/56 =

SEES

SESE

SSEE

ESSS And so on, but...a big BUT...why do all this when

SESS

SSES

SSSE

SSSS

p(E>=1|4) = 1 - p(S and S and S and S) = 1 - 24/59 * 23/58 * 22/57 * 21/56 = 0.976652619 ANS. In other words we find the probability of not meeting an Englishman and take 1 minus that value to find the probability of meeting at least one.

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Step-by-step explanation:

3 0

2 years ago

In the United States, 30% of people own a cat and 3% of people own a bird. What is the probability of someone owning both a cat

Alex777 [14]

Answer: 9/100

Step-by-step explanation:

Probability of people own a cat:

P(cat)=30/100=0.3

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P(bird)=3/100=0.03

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3 years ago

A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected

Fittoniya [83]

Answer:

(a) The probability that the members of the committee are chosen from all nationalities $=\frac{4}{33}$ =0.1212.

(b)The probability that all nationalities except Italian are represent is 0.04848.

Step-by-step explanation:

Hypergeometric Distribution:

Let $x_1$ , $x_2$ , $x_3$ and $x_4$ be four given positive integers and let $x_1+x_2+x_3+x_4= N$ .

A random variable X is said to have hypergeometric distribution with parameter $x_1$ , $x_2$ , $x_3$ , $x_4$ and n.

The probability mass function

$f(x_1,x_2.x_3,x_4;a_1,a_2,a_3,a_4;N,n)=\frac{\left(\begin{array}{c}x_1\\a_1\end{array}\right)\left(\begin{array}{c}x_2\\a_2\end{array}\right) \left(\begin{array}{c}x_3\\a_3\end{array}\right) \left(\begin{array}{c}x_4\\a_4\end{array}\right) }{\left(\begin{array}{c}N\\n\end{array}\right) }$

Here $a_1+a_2+a_3+a_4=n$

${\left(\begin{array}{c}x_1\\a_1\end{array}\right)=^{x_1}C_{a_1}= \frac{x_1!}{a_1!(x_1-a_1)!}$

Given that, a foreign club is made of 2 Canadian members, 3 Japanese members, 5 Italian members and 2 Germans members.

$x_1$ =2, $x_2$ =3, $x_3$ =5 and $x_4$ =2.

A committee is made of 4 member.

N=4

(a)

We need to find out the probability that the members of the committee are chosen from all nationalities.

$a_1$ =1, $a_2$ =1, $a_3$ =1 , $a_4$ =1, n=4

The required probability is

$=\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\1\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{2\times 3\times 5\times 2}{495}$

$=\frac{4}{33}$

=0.1212

(b)

Now we find out the probability that all nationalities except Italian.

So, we need to find out,

$P(a_1=2,a_2=1,a_3=0,a_4=1)+P(a_1=1,a_2=2,a_3=0,a_4=1)+P(a_1=1,a_2=1,a_3=0,a_4=2)$

$=\frac{\left(\begin{array}{c}2\\2\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\2\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$ $+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\2\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{1\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 1}{495}$

$=\frac{6+12+6}{495}$

$=\frac{8}{165}$

=0.04848

The probability that all nationalities except Italian are represent is 0.04848.

6 0

3 years ago