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3 years ago

How do I solve 12x^2-89x-56 factor by grouping

1 answer:

7 0

Use ac method

multiply 12 and -56
-672
now find what 2 factors of -672 add to -89
answer is -96 and 7

expand the middle term like that
12x²-96x+7x-56
group
(12x²-96x)+(7x-56)
factor
12x(x-8)+7(x-8)
undistribute the (x-8)
(x-8)(12x+7) is answer

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2 years ago

A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat

Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =16$ represent the sample variance

s=4 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=36.191$

$\chi^2_{1- \alpha/2}=7.633$

And replacing into the formula for the interval we got:

$\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}$

$8.400 \leq \sigma^2 \leq 39.827$

So the 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

7 0

3 years ago

What is the force of gravity (in Newtons) acting between the Earth and a 100-kg person who is on board the International Space S

ratelena [41]

Answer:

$f_{G}=317877.55 N = 317.8 kN$

Step-by-step explanation:

$f_{G}= (G*m_{1}*m_{2})/r^{2} \\Where:\\f_{G}= force of gravity\\G=gravitational constant= 6.6x10^{-11}m^{3}kg^{-1}s^{-2}\\ m_{1}= earth mass = 5.9x10^{24}kg\\m_{2}=person mass = 100kg\\r= distance between the bodies=350km= 350 000 m$

5 0

2 years ago

Given that f(x)=5x^3+ax^2-4x-2 and that (x-1) is a factor of f(x), find a.

stepladder [879]

Answer:

a=1

Step-by-step explanation:

Because x-1 is factor of f(x)

for x1=1, f(1)=0, then

f(1)=5*1^3+a*1^2-4*1-2=0

5*1+a*1-4-2=0

5+a-6=0

a-1=0

a=1

5 0

3 years ago