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Elan Coil [88]
2 years ago
8

BRAInly whats the difference between isolation and separation?

Mathematics
1 answer:
Nina [5.8K]2 years ago
8 0

is that separation is the act of separating or the condition of being separated while isolation is (chiefly|uncountable) the state of being isolated, detached, or separated.

also, whats up with the singing?

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Identify the variables in this Algebraic Expression<br> X+ 4y - 9
Lesechka [4]

Answer:

The expression x + 4y - 9 has two variables x and y.

Step-by-step explanation:

<u>What is a variable?</u>

A variable is a letter, for instance x or y, that represents an unknown number.

Given the expression

x + 4y - 9

The given expression has three terms which are x, 4y, and -9.

Here:

  • The term 'x' is the unknown variable x.
  • The term '4y' has a coefficient (the number next to a variable y) of 4.
  • The term '-6' is a constant.

As we already know that a variable is a letter, for instance x or y, that represents an unknown number. In other words, the values of x, or y are unknown to us.

Thus, the expression x + 4y - 9 has two variables x and y.

5 0
3 years ago
A small airplane can hold 44 passengers. Fifteen passengers board the plane. what is the solution.
Arada [10]
If I understand this problem correctly the answer would be 29 more people can board the plane! Hope this helps! Comment if you have questions!!!
6 0
3 years ago
Help me please ASAP!!!!!!
FrozenT [24]

Answer:

45 degrees

Step-by-step explanation:

Ok I think it is 45 degrees try that if not super sorry

3 0
2 years ago
HELP ASAP 10 POINTS TO BRAINLIEST
Brilliant_brown [7]

Answer:

15%

Step-by-step explanation:

The cost of 2 adults and 2 children without the discount is

2 adults = 2 (55) = 110

2 children = 2 (45) = 90

Total cost = 110+90 = 200

The original price is 200 and the new price is 170

Percentage discount = (Original price - new price)/ original price * 100%

                                     = (200-170)/200 * 100%

                                     = 30/200 * 100%

                                     = .15 * 100%

                                     = 15%

6 0
3 years ago
Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1
alexandr1967 [171]

Answer:

a) \simeq 0.3012   b) \simeq 0.0494 c) \simeq 0.2438

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, \frac{1.2}{4}

= 0.3 collisions per  month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

          P(X =x) = \frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}&#10;

                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, \lambda = 0.3 collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

         P(X = 0) = \frac{e^{-0.3}\times {\0.3}^{0}}{0!}&#10;

                      \simeq 0.7408182207 ------------------------------------(2)

Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =0.7408182207^{4}

     \simeq 0.3012  -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

           P(X =1) = \frac{e^{-0.3}\times {\0.3}^{1}}{1!}&#10;

                      \simeq 0.2222454662 ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

                =0.2222454662^{2}

                \simeq 0.0494 -----------------------------------------(5)

1 collision in 6 months period means

                                \frac{1}{6} collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6]  (which is to be estimated)

=\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

\simeq 0.6543894283-------------------------------------------(6)

So,

P(1  collision in 6 month period)

  =  0.6543894283^{6}

   \simeq 0.0785267444 ------------------------------------------------(7)

So,

P(No collision in 6 months period)

  = (P(X =0)^{6}

   \simeq 0.1652988882 ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

\simeq  0.2438 ---------------------------------------------(9)          

7 0
2 years ago
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