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juin [17]
3 years ago
9

Can someone plz help me on this I really need help

Mathematics
1 answer:
Usimov [2.4K]3 years ago
5 0

When dividing fractions these are the steps you will take:

1. The first number in the expression stays the same. If it is a whole number then you may just place a one in the denominator and keep the numerator as the whole number like so:

\frac{3}{1} ÷ \frac{3}{4}

2. Change the division sign into a multiplication sign

\frac{3}{1} × \frac{3}{4}

3. Take the reciprocal (switch the places of numerator and denominator) of the second number in the expression

\frac{3}{1} × \frac{4}{3}

4. Multiply across

\frac{3*4}{1*3}

\frac{12}{3}

^^^This can be further simplified:

4

Hope this helped!

~Just a girl in love with Shawn Mendes

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Answer:

Distance of the trail from the start to the end is 6\frac{17}{24} miles

Step-by-step explanation:

Distance from the start of a trail to the bird lookout point=3\frac{7}{8} miles

Distance from the bird lookout point to the end of the trail =2\frac{5}{6} miles

We want to find the difference of the trail from the start to the end, that is, how far is the starting point from the end point.

The distance between the start and end points

=3\frac{7}{8}+2\frac{5}{6}\\=\frac{31}{8} +\frac{17}{6}

=\frac{161}{24} =6\frac{17}{24} miles

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In a group of 10 ​kittens, 5 are female. Two kittens are chosen at random. ​a) Create a probability model for the number of male
olya-2409 [2.1K]

Answer:

(b) Expected number of male kitten E(x) = 1

(c) Standard deviation is ±\frac{2}{3}.

Step-by-step explanation:

Given that,

Number of kittens in a group are 10. Out of these 5 are female.

To find:- (a) create a probability model of male kitten chosen.

    So,  total number of kitten = 10

           number of female kitten = 5    

then, Number of male kitten = 10-5= 5

Now total number of ways to choosing two kittens from group of 10 = ^{10} C_{2}

                                                                                                                 = \frac{10!}{2!\times8!}

                                                                                                                 = 45

for choosing (i) No male P(x=0) = P(Two female) =  \frac{^{5} C_{2}}{45}

                                                                           = \frac{2}{9}

                     (ii) 1 male P(x=1) = P(1 male and 1 female) = \frac{^{5} C_{1}\times^{5} C_{1}}{45} =  \frac{25}{45} =\frac{5}{9}

                     (iii) 2 male P(x=2)= P(2 male) =\frac{^{5} C_{2}}{45} =  \frac{2}{9}

      x_{i}                             0                     1                      2

P(X=x_{i})                           \frac{2}{9}                      \frac{5}{9}                      \frac{2}{9}

(b) what is expected number of male kittens chosen ?

    Expected number of male kitten E(x) =  o\times \frac{2}{9} + 1\times \frac{5}{9} + 2\times\frac{2}{9}

                                                                  = 1

(c) what is standard deviation ?

                              \sigma(x) = \sqrt{ (E(x^{2} )-(Ex)^{2})

No,                          

                               Ex^{2} =o\times \frac{2}{9} + 1\times \frac{5}{9} + 4\times\frac{2}{9}

                                       = \frac{13}{9}

                               \sigma(x)=\sqrt{\frac{13}{9} - 1 }=\sqrt{\frac{4}{9} }  

                                       = ± \frac{2}{3}

                               

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