Answer:
Step-by-step explanation:
8.) For a triangles sides to make sense, you must be able to add up two values of the triangle, and the result should be more than the third side. Add the lowest values and see if the result is greater than the biggest number:
![4.2+7.9=12.1](https://tex.z-dn.net/?f=4.2%2B7.9%3D12.1)
12.1 is less than the given side, 13.3, so a triangle cannot have the lengths.
10.) 6<x<22
To find the range for the third side of the triangle, you need to find how small x can be (the missing side) and you need to see how large it can be.
You need to see how small it can be because any two sides have to be greater than the third side. You also need to see how big it can be because, if it's too big, the other two sides will be less than the third side, which would make an open shape (see picture).
To find the range, first see how small. Subtract the known sides:
![22-6=16](https://tex.z-dn.net/?f=22-6%3D16)
So, x has to be greater than 16.
x > 16
Now add the known sides:
![22+6=28](https://tex.z-dn.net/?f=22%2B6%3D28)
x needs to be less than 28 for the other two sides to be greater than x:
x < 28
Insert the inequalities into a single inequality:
16 < x <28
X has to be greater than x, but less than 28.
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}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumber%20%5Chspace%7B0.1cm%7D%20of%20%5Chspace%7B0.1cm%7D%20secreataries%7D%7Btotal%20%5Chspace%7B0.1cm%7D%20number%20%5Chspace%7B0.1cm%7D%20of%20%5Chspace%7B0.1cm%7D%20employees%7D%20%20%3D%20%5Cfrac%7B6%7D%7B15%7D)
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)}](https://tex.z-dn.net/?f=%5Cfrac%7Bp%28%28consultant%20%5Chspace%7B0.1cm%7D%20wins%20%5Chspace%7B0.1cm%7D%20on%20%5Chspace%7B0.1cm%7D%20first%20%5Chspace%7B0.1cm%7Ddraw%29%5Ccap%28%20secretary%5Chspace%7B0.1cm%7D%20wins%5Chspace%7B0.1cm%7D%20on%20second%20%5Chspace%7B0.1cm%7Ddraw%29%29%7D%7Bp%28consultant%20%5Chspace%7B0.1cm%7D%20wins%20%5Chspace%7B0.1cm%7D%20on%20%5Chspace%7B0.1cm%7D%20first%20%5Chspace%7B0.1cm%7Ddraw%29%7D)
=
.
The probability that a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw =
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}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumber%20%5Chspace%7B0.1cm%7D%20of%20%20%5Chspace%7B0.1cm%7D%20secretaries%20%20%5Chspace%7B0.1cm%7D%20remaining%20%7D%20%7B%20number%20%20%5Chspace%7B0.1cm%7D%20of%20%20%5Chspace%7B0.1cm%7D%20employees%20%20%5Chspace%7B0.1cm%7D%20remaining%7D%20%20%3D%20%5Cfrac%7B6%20-%201%7D%7B15%20-%201%7D%20%20%3D%20%5Cfrac%7B5%7D%7B14%7D)
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}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumber%20%5Chspace%7B0.1cm%7D%20of%20%20%5Chspace%7B0.1cm%7D%20consultants%20%20%5Chspace%7B0.1cm%7D%20%20%7D%20%7B%20number%20%20%5Chspace%7B0.1cm%7D%20of%20%20%5Chspace%7B0.1cm%7D%20employees%20%20%5Chspace%7B0.1cm%7D%20%7D%20%20%3D%20%5Cfrac%7B5%20%7D%7B15%7D%20%20%3D%20%5Cfrac%7B1%7D%7B3%7D)
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}](https://tex.z-dn.net/?f=%5Cfrac%7B6%7D%7B15%7D%20%20%5Ctimes%20%5Cfrac%7B5%7D%7B14%7D%20%20%3D%20%5Cfrac%7B1%7D%7B7%7D)
Therefore the only statement that is not true is b.)
Answer:
5.24*10³= 5.24*1000= 5240 as decimal shifts towards right
524000/10²=524000/100=5240
Step-by-step explanation:
Answer:
3×(27-7) says to subtract 7 from 27 first, then multiply with 3; but 3×27-7 says to multiply 3 by 27 first ,then subtract 7.
The second bullet is the correct option