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tatuchka [14]
3 years ago
5

Four small cakes are shared equally among 5 children what part does each child receive?

Mathematics
1 answer:
erastova [34]3 years ago
4 0

Answer:

0.8 of a cake

Step-by-step explanation:

all you do is divide 4 by 5

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Simplify the product (x-4)(x+3)
Kamila [148]

Answer:

Final answer is x^2-x-12.

Step-by-step explanation:

Given expression is \left(x-4\right)\left(x+3\right).

Now we need to find the product of the given expression \left(x-4\right)\left(x+3\right).

\left(x-4\right)\left(x+3\right)

=x\left(x-4\right)+3\left(x-4\right)

=x^2-4x+3x-12

combine like terms

=x^2-1x-12

=x^2-x-12

Hence final answer is x^2-x-12.

7 0
4 years ago
Read 2 more answers
Simplify. y^-3 pls help me i need this very bad
prisoha [69]
My understanding is that when the power is a negative number, move it to the denominator and change it to positive. 
In this case, the base is y, the power is -3
y^(-3)=1/(y^3)
7 0
3 years ago
PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!
ollegr [7]

Answer: For 1 it is A - f(x)=(x-2)(x+4) and for 2 it is C - y = 0.5x^2 + 1.5x − 5

Step-by-step explanation:

Just use Desmos

7 0
1 year ago
Question:
faltersainse [42]

Answer:

The probability that a ship that is declared defecive is sound is 0.375

Step-by-step explanation:

Let P(A|B) denote the conditional probability of A given B. We will make use of the equation

P(A|B) = P(A) × P(B|A) / P(B)

We have the probabilities:

  • P(Declared Defective (detected) | Defective) = 0.95
  • P(not Detected | Defective) = 1-0.95=0.05
  • P(Declared Sound | Sound) = 0.97
  • P(Declared Defective |Sound) = 1-0.97=0.03
  • P(Defective)=0.05
  • P(Sound)= 1- 0.05 = 0.95

We can calculate:

P(Declared Defective)= P(Detected | Defective)×P(Defective) + P(Declared Defective |Sound) ×P(Sound) = 0.95×0.05 + 0.03×0.95=0.076

P(S | Declared Defective) =

(P(Sound) × P(Declared Defective | Sound)) / P(Declared Defective)

=0.95×0.03 /0.076 =0.375

3 0
3 years ago
Can someone help me please
gizmo_the_mogwai [7]
The answer is 25 chicks

Reasoning:

This is a simple problem.
Consider you are the only chick that matters, and construct a table to say whether YOU get pecked. Your chance of being pecked comes down to only 4 outcomes. (1) YES - pecked twice. (2) YES - pecked from left wing only. (3) YES - pecked from right wing only. (4) NO - unpecked.
The table has 4 elements, all of equal probability, 1 of which is unpecked. YOU are therefore pecked 3:1 ratio or 3:4 opportunities  75% of the time. For convenience, this needs to be conducted for 100 trials of YOU, and the answer is that 25 times YOU will NOT be pecked. The circular nature of the 100 chicks says that YOU are not unique, and your experience is the same as the others, so we extrapolate your experience of 100 trials to a single trial of 100 chicks just like YOU. 25 unpecked chicks, 50 get pecked once, 25 get double pecks.

This is the same table constructed for 100 women having two children and asking how many have no girls.
7 0
3 years ago
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