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3 years ago

Factor the polynomial expression x4 + 18x2 + 81

Mathematics

1 answer:

Akimi4 [234]3 years ago

4 0

Answer: $(x^2+9)^2$

Step-by-step explanation:

Rewrite it in the form $a^2+2ab+b^2$ , where $a=x^2 and b=9.$

$(x^2)^2+2(x^2)(9)+9^2$

Use Square of Sum: $(a+b)^2= a^2+2ab+b^2$

$(x^2+9)^2$

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Consider the set a = {a, b, {a, b}}, where a and b are distinct elements. the number of elements in ℘(a) is _____

Juli2301 [7.4K]

$A=\{a,b,\{a,b\}\}\\|A|=3\\\\|\wp(A)|=2^3=8$

7 0

3 years ago

Suppose that a standardized biology exam has a mean score of 80% correct, with a standard deviation of 3. The school administrat

NemiM [27]

Answer:

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

A sample of 65 students from the freshmen class is used and a mean score of 76% correct is obtained.

This means that $n = 65, \pi = 0.76$

99% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 - 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.6236$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 + 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.8964$

0.6236*100 = 62.36%

0.8964*100 = 89.64%

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).

6 0

3 years ago

Solving for matrices

muminat

Answer:

Step-by-step explanation:

The augmented matrix for the system of three equaitons is

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\5&2&-2&|&11\\5&-4&4&|&-7\end{array}\right)$

Multiply the first row by 5, the second row by -3 and add these two rows:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\5&-4&4&|&-7\end{array}\right)$

Subtract the third row from the second:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&6&-6&|&18\end{array}\right)$

Divide the third row by 6:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&1&-1&|&3\end{array}\right)$

Now multiply the third equation by 26 and add it to the second row:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&0&-45&|&-90\end{array}\right)$

You get the system of three equations:

$\left\{\begin{array}{r}3x-4y-5z=-27\\-26y-19z=-168\\-45z=-90\end{array}\right.$

From the third equation

$z=\dfrac{90}{45}=2.$

Substitute z=2 into the second equation:

$-26y-19\cdot 2=-168\\ \\-26y-38=-168\\ \\-26y=-168+38=-130\\ \\y=\dfrac{130}{26}=5.$

Now substitute z=2 and y=5 into the first equation:

$3x-4\cdot 5-5\cdot 2=-27\\ \\3x-20-10=-27\\ \\3x-30=-27\\ \\3x=-27+30=3\\ \\x=1.$

The solution is (1,5,2)

4 0

3 years ago

He answer to all of the questions please

AlladinOne [14]

I think it's C,But I'm not sure

5 0

4 years ago

Customers arrive at a grocery store at an average of 2.1 per minute. Assume that the number of arrivals in a minute follows the

Black_prince [1.1K]

Answer:

a) P(2)=0.270

b) P(X>3)=0.605

c) P=0.410

Step-by-step explanation:

We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the Poisson distribution:

$\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}$

a) In this case: $\lambda=2.1$

$P(2)=\frac{2.1^2 \cdot e^{-2.1}}{2}\\\\P(2)=0.270$

Therefore, the probability is P(2)=0.270.

b) In this case: $\lambda=2\cdot 2.1=4.2$

$P(X>3)=1-P(X\leq 3)\\\\P(X>3)=1-\sum_{x=0}^3 \frac{4.2^x \cdot e^{-4.2}}{x!}\\\\P(X>3)=1-0.395\\\\P(X>3)=0.605$

Therefore, the probability is P(X>3)=0.605.

c) We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.

In this case: $\lambda=2\cdot 2.1=4.2$

$P(X\geq 5)=1-P(X$

Therefore, the probability is P=0.410.

4 0

3 years ago