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slava [35]
3 years ago
5

What is the factored form of 8X^2 +12X?

Mathematics
1 answer:
pogonyaev3 years ago
8 0
The highest common factor of both terms is 4x, so take this out by dividing each term by 4x

8x^2 / 4x = 2x
12x / 4x = 3

So the factored form is:
4x(2x+3)
You might be interested in
PLEASE HELP!
Marina86 [1]

Answer:

(x + 2) + (–5x + 4) / (x² – 2x + 1)

Step-by-step explanation:

(x³ – 8 x + 6) ÷ (x² – 2x + 1)

The operation can be carried out as follow:

Please see attached photo for details.

(x³ – 8 x + 6) ÷ (x² – 2x + 1) =

(x + 2) + (–5x + 4) / (x² – 2x + 1)

4 0
3 years ago
Draw a conclusion is the statement below always sometimes or never true give at least two examples to support your reasoning the
hodyreva [135]
It's sometimes true.

One example is the least common multiple of 2 and 3 is 6, which is their product.

But the product isn't always the answer because (example 2:) the least common multiple of 6 and 10 is 30 because 6*5=30 and 3*10=30, however 6*10 is 60.

Ergo, it is only sometimes true.
3 0
3 years ago
The volume of a cone varies jointly with the area of the base and the height. When the area of the base is 27 cm2 and the height
Tomtit [17]
Cone=cup:
V=124<span>cm^3.
h=12cm
V=Bh
P=r</span>^2π+rsπ:
B=r^2π
B=V/h=124/12=10,3cm^2
r^2=B/π
r=√(B/π)=1,81 cm
S=√(r^2+h^2)=√(12^2+1.81^2)=√(144+3,28)=12,13 cm
P=r^2π+rSπ=3,28*3,14+1,81*12,13*3,14=10,29+68,93=79,22 cm^2
3 0
3 years ago
Best known for its testing program, ACT, Inc., also compiles data on a variety of issues in education. In 2004 the company repor
GarryVolchara [31]

Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

5 0
3 years ago
Complete the following direct proof:
anzhelika [568]
B. I hope this help please let me know if it's correct.
4 0
3 years ago
Read 2 more answers
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