I don’t know how to do this and my teacher doesn’t know how to teacher please help me

A set of equations is shown above. Which method eliminates one of the variables? A) Multiply equation A by -1/3 and add the resu

Paraphin [41]

Answer:

Step-by-step explanation:

Multiplying Equation A by (1/3) and adding the result to Equation B will do the trick. Let's actually solve the problem!

Equation A: (5/3)x + 3y = 12

Equation B: 4x - 3y = 8

---------------------------

(5/3 + 12/3)x = 15 Note how this has eliminated the variable

(17/3)x = 15 y.

x = (3/17)(15)

8 0

3 years ago

I need to an explanation about this formula.

sergeinik [125]

I would answer but i can not see what it says.... Sorry

6 0

3 years ago

. Lightning Strikes It has been said that the probability of being struck by lightning is about 1 in 750,000, but under what cir

mrs_skeptik [129]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$P(U |D ) = 0.198$

b

$P(O\ n \ B) = 0.188$

c

$P(O | B) = 0.498$

Step-by-step explanation:

The total number of deaths is mathematically represented as

$T = 16 + 23 + \cdots + 16$

$T =623$

The total number of deaths in 1996 - 2000 is mathematically represented as

$T_a = 16 + 23+ \cdots + 30$

$T_a = 235$

The total number of deaths in 2001 - 2005 is mathematically represented as

$T_b = 17 + 16 + \cdots + 23$

$T_b = 206$

The total number of deaths in 2006 - 2010 is mathematically represented as

$T_c = 15 + 17 + \cdots + 16$

$T_d = 182$

Generally the the probability that it would occur under the tree given that the death was after 2000 is mathematically represented as

$P(U |D ) = \frac{P(A \ n\ U )}{P(A)}$

Here $P(A \ n\ U )$ represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

$P(A \ n\ U ) = \frac{Z}{ T}$

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

$Z = 35 + 42$

=> $Z = 77$

=> $P(A \ n\ U ) = \frac{77}{ 623}$

=>

Also

$P(A)$ is the probability of the death occurring after 2000 and this is mathematically represented as

$P(A) = \frac{T_b + T_c}{ T}$

=> $P(A) = \frac{ 206+ 182}{623}$

=>

So

$P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+ 182}{623}}$

=> $P(U |D ) = 0.198$

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

$P(O | B) = \frac{T_z}{ T}$

Here $T_z$ is the total number of death outside / camping before 2001 and the value is 117

So

$P(O \ n \ B) = \frac{117}{623}$

=> $P(O\ n \ B) = 0.188$

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

$P(O | B) = \frac{ P( O \ n \ B)}{ P(B)}$

Here $P(B)$ is the probability that it was before 2001 , this is mathematically represented as

$P(B ) = \frac{T_a}{T}$

=> $P(B ) = \frac{235}{623}$

So

$P(O | B) = \frac{ \frac{117}{623}}{ \frac{235}{623}}$

=> $P(O | B) = 0.498$

5 0

3 years ago

How do we know what sign goes outside of the bracket when factoring binomials?

Xelga [282]

I'm assuming you're asking about negative or positive signs here. When I factor binomials, it depends on what would make factoring simpler for me. And how you're factoring the binomial too.

More often than not, it's going to be positive. For example, 4x^2 - 2x = 0. That would factor out to be 2x(2x-1) = 0. The sign outside is going to be positive because it's simpler to solve that way. Here, x is equal to 0 and equal to 1/2. Taking out a negative in this case would complicate the solving process.

3 0

3 years ago

A halloween assorted bag of candy contains 4 Snickers, 2 packages of M&M's, & 3 Butterfingers. What is the probability o

a_sh-v [17]

Answer:

2/27

Step-by-step explanation:

In the question we have:

4 snickers

2 packages of M&M's

3 Butterfingers.

Total number of candies in the Halloween bag = 9

Probability of selecting some M&M's and putting them back = 2/9

Probability of selecting a Butter finger = 3/9

Therefore, the probability of selecting some M&M's, putting them back, and then selecting a Butterfinger =

Probability of selecting some M&M's and putting them back × Probability of selecting a Butter finger

2/9 × 3/9

= 6/ 81

= 2/27

3 0

3 years ago