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aleksandr82 [10.1K]
3 years ago
8

Write a cubic trinomial that has 3x as a GCF of its terms.

Mathematics
2 answers:
vichka [17]3 years ago
7 0
(3x + 9x + 12x^2) hope this helps!
SpyIntel [72]3 years ago
3 0
Ax^3 + Bx+^2 + Cx is the correct answer
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156, 153, 150, ...<br> Find the 30th term.
Maksim231197 [3]

Answer:

First multiply 30 by 3 = 90 then subtract 90 from 156 to get 66

Step-by-step explanation:

66 would be your thirtieth term.

3 0
4 years ago
Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. These data are contained in the
galina1969 [7]

Answer:

a. The margin of error is 2.29.

b. 19.23 to 23.81

Step-by-step explanation:

The sample size is n=64.

We start by calculating the sample mean and standard deviation with the following formulas:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}

The sample mean is M=21.52.

The sample standard deviation is s=6.89.

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{6.89}{\sqrt{64}}=\dfrac{6.89}{8}=0.861

The degrees of freedom for this sample size are:

df=n-1=64-1=63

The t-value for a 99% confidence interval and 63 degrees of freedom is t=2.656.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.656 \cdot 0.861=2.29

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 21.52-2.29=19.23\\\\UL=M+t \cdot s_M = 21.52+2.29=23.81

The 99% confidence interval for the mean is (19.23, 23.81).

5 0
3 years ago
Please help!!!!!!!!!!!¡
Daniel [21]
Your answer would be ( -12/5 x + 2 )
6 0
3 years ago
What is the value of s in the equation below? S + 1 1/2 = 10 1/2
LenaWriter [7]

Answer:

S=9

Step-by-step explanation:

10\frac{1}{2} - 1\frac{1}{2} = 9

6 0
3 years ago
Read 2 more answers
Choose whether it's always, sometimes, never 
Keith_Richards [23]

Answer: An integer added to an integer is an integer, this statement is always true. A polynomial subtracted from a polynomial is a polynomial, this statement is always true. A polynomial divided by a polynomial is a polynomial, this statement is sometimes true. A polynomial multiplied by a polynomial is a polynomial, this statement is always true.

Explanation:

1)

The closure property of integer states that the addition, subtraction and multiplication is integers is always an integer.

If a\in Z\text{ and }b\in Z, then a+b\in Z.

Therefore, an integer added to an integer is an integer, this statement is always true.

2)

A polynomial is in the form of,

p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0

Where a_n,a_{n-1},...,a_1,a_0 are constant coefficient.

When we subtract the two polynomial then the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3)

If a polynomial divided by a polynomial  then it may or may not be a polynomial.

If the degree of numerator polynomial is higher than the degree of denominator polynomial then it may be a polynomial.

For example:

f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2

Then \frac{f(x)}{g(x)}=x^2+5, which a polynomial.

If the degree of numerator polynomial is less than the degree of denominator polynomial then it is a rational function.

For example:

f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2

Then \frac{g(x)}{f(x)}=\frac{1}{x^2+5}, which a not a polynomial.

Therefore, a polynomial divided by a polynomial is a polynomial, this statement is sometimes true.

4)

As we know a polynomial is in the form of,

p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0

Where a_n,a_{n-1},...,a_1,a_0 are constant coefficient.

When we multiply the two polynomial, the degree of the resultand function is addition of degree of both polyminals and the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3 0
3 years ago
Read 2 more answers
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