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Varvara68 [4.7K]
3 years ago
8

What is the distance between the points -1,-3) and (-1,43)? I think I'll die if I don't get the answer soon​

Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
6 0
46.

Ther are both at the same x-coordinate, so 43-(-3)=46
Phoenix [80]3 years ago
4 0

Answer:

you mean the midpoint: (-1, 23)

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13. If the following fractions were converted to decimals, which one would result in a repeating decimal? A. 3/4 B. 1/9 C. 5/11
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Hey there!

In order to find if a fraction would result in a repeating decimal, recall that a fraction is a division problem written vertically. All that you have to do is divide the numerator by the denominator. Also, remember that a repeating decimal will result in the same number after the decimal point as long as the calculator can handle. 

3 ÷ 4 = 0.75

1 ÷ 9 = 0.11111111...

5 ÷ 11 = 0.45454545...

3 ÷ 0.42857143...

As you can see, two out of your four answer choices give you a repeating decimal. B gives you a repeated number of "1" while C gives you "45". D doesn't count since there is no pattern of repeated numbers that it follows. 

Both B and C fall into the category of repeating decimal. If you're only able to choose one answer, I would ask your teacher, a parent, or a peer what they think. 

Hope this helped you out! :-)
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A shop buys pairs of trainers for £20 each. They sell them to make a 30% profit on each pair. What is the selling price of the t
Alexxx [7]

Answer:

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Step-by-step explanation:

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What is 1755 divide by 27
Zielflug [23.3K]
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Find (a) the circumference and (b) the area of the circle. Use 3.14 or 22/7 for pi. Round your answer to the nearest whole numbe
balandron [24]

Answer:

a)219.8

b)3846.5

Step-by-step explanation:

a) circumference is pi*diameter, so 3.14*70 is 219.8

b) area is pir^2, and radius is 1/2 of diameter, it would be 3.14*35^2, so 3.14*1225, so 3,846.5 in^2.

4 0
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Read 2 more answers
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
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