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antiseptic1488 [7]
3 years ago
14

I need help on these two questions please.

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
7 0

Answer:

They are similar

Step-by-step explanation:

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At the start of 2014 Tim's house was worth 100,000.
Novay_Z [31]
For this case we have a function of the form:
 y = A * (b) ^ t
 Where,
 A: initial amount
 b: growth rate
 t: time
 Substituting values we have:
 y = 100000 * (1.10) ^ 4
 y = 146410
 Answer:
 
the value of his house at the start of 2018 is:
 
y = 146410
5 0
4 years ago
Please Please Please Help Me
maks197457 [2]

Answer:

  1. (x-y^2)(x^2 +xy^2 +y^4)
  2. (a^2 +b)(a^4 -a^2b +b^2)
  3. (m^3-n)(m^6 +m^3n +n^2)
  4. (p+k^3)(p^2 -pk^3 +k^6)
  5. (a^2+b^3)(a^4 -a^2b^3 +b^6)
  6. (x-y)(x^2 +xy +y^2)(x^6 +x^3y^3 +y^6)

Step-by-step explanation:

In every case, the factorization makes use of the standard form for factoring the sum or difference of cubes:

  • a^3 +b^3 = (a +b)(a^2 -ab +b^2)
  • a^3 -b^3 = (a -b)(a^2 +ab +b^2)

1. a=x, b=y^2. Use the formula for the difference.

2. a^2 ⇒ a, b = b. Use the formula for the sum.

3. a=m^3, b=n. Use the formula for the difference.

4. a=p b=k^3. Use the formula for the sum.

5. a^2 ⇒ a, b^3 ⇒ b. Use the formula for the sum.

6. a=x^3, b=y^3. Use the formula for the difference. When you do, the first factor is the difference x^3 -y^3, which can be factored using the difference formula again with a=x, b=y.

4 0
4 years ago
Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, on
Iteru [2.4K]

Answer:

P(X_i=2) =\dfrac{1}{6}

P(X_i=-1) =\dfrac{5}{6}

Step-by-step explanation:

Given the numbers on the chips = 1, 1, 3 and 5

Miguel chooses two chips.

Condition of winning: Both the chips are same i.e. 1 and 1 are chosen.

Miguel gets $2 on winning and loses $1 on getting different numbers.

To find:

Probability of winning $2 and losing $1 respectively.

Solution:

Here, we are given 4 numbers 1, 1, 3 and 5 out of which 2 numbers are to be chosen.

This is a simple selection problem.

The total number of ways of selecting r numbers from n is given as:

_nC_r = \frac{n!}{r!(n-r)!}

Here, n = 4 and r = 2.

So, total number of ways = _4C_2  = \frac{4!}{2!\times 2!} = 6

Total number of favorable cases in winning = choosing two 1's from two 1's i.e. _2C_2 = \frac{2!}{2! 0! } = 1

Now, let us have a look at the formula of probability of an event E:

P(E) = \dfrac{\text{Number of favorable ways}}{\text{Total number of ways}}

So, the probability of winning.

P(X_i=2) =\dfrac{1}{6}

Total number of favorable cases for -1: (6-1) = 5

So, probability of getting -1:

P(X_i=-1) =\dfrac{5}{6}

Please refer to the attached image for answer table.

7 0
3 years ago
8/10 = w/15 what does w stand for
nydimaria [60]

Answer:

Step-by-step explanation:

its just an unknown variable usually it is one

3 0
3 years ago
Read 2 more answers
22,602,188,507 what is the value of the digit 6 in the number
Artist 52 [7]

the value of 6 is 6 hundred million

4 0
3 years ago
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