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nadezda [96]
3 years ago
9

Solve for x 6x + 11= - (6x + 5)

Mathematics
2 answers:
vitfil [10]3 years ago
6 0

Answer:

x = -4/3

Step-by-step explanation:

6x + 11= - (6x + 5)

Distribute the minus sign

6x+11 = -6x-5

Add 6x to each side

6x+11 +6x = -6x-5+6x

12x+11 = -5

Subtract 11 from each side

12x+11-11 = -5-11

12x = -16

Divide by 12

12x/12 = -16/12

x = -4/3

Semmy [17]3 years ago
3 0

Answer:

x = -4/3

Step-by-step explanation:

6x + 11 = -6x - 5

12x = -16

x = -4/3

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3x^2-9x+0<br> Factor the trinomial
sineoko [7]

Answer:

3x (x-3)

Step-by-step explanation

3x^2 - 9x + 0

The equation.

3x^2 - 3^2 × x

What I did here is that I factored out 9x to 3^2 times x.

3x (3x^2 all over 3x - 3^2 × x/ all over 3x)

So that way I can divide the number in the parentheses by 3x to take 3x out of the parentheses to factor out.

3x (x^2-1 - (3^2-1) )

You simplify to get the final answer.

3x (x-3)

Final Answer.

I hope this helps!!!!

8 0
3 years ago
A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents di
cupoosta [38]

Answer:

The probability value is almost equal to 0. Implying that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of decedents who died in the three-month period following their birthdays.

A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents died in the three-month period following their birthdays (123).

The probability (p) of anyone dying in any quarter if people die randomly during the year is simply 0.25.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 747 and p = 0.25.

But the sample selected is too large and the probability of success is small.

So a Normal approximation to binomial can be applied to approximate the distribution of <em>p</em> if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=747\times 0.25=186.75>10\\n(1-p)=747\times (1-0.46)=560.25>10

Thus, a Normal approximation to binomial can be applied.

So,  p\sim N(\hat p,\ \frac{\hat p(1-\hat p)}{n}).

Compute the probability that 46% or more would die in that particular interval if deaths occurred randomly throughout the year as follows:

P (p\geq 0.46)=P(\frac{p-\hat p}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}>\frac{0.46-0.25}{\sqrt{\frac{0.25(1-0.25)}{747}}})

                   =P(Z>13.25)\\=1-P(Z

 *Use a <em>z</em> table for the probability.

The probability value is almost equal to 0. This probability is very low indicating that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

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3 years ago
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Answer:

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Step-by-step explanation:

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Last year, the Humane Society of a certain city took in 21,320 animals. If the Humane Society was open all 52 weeks that year, h
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21,320 animals / 52 weeks
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Answer:

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Step-by-step explanation:

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4 years ago
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