Holly says 4.131131113... is a rational number. Which of the following best describes whether Holly is correct and why?

What is the volume of papers that the tray can hold?

Vesna [10]

What tray..there is not enough information to help you

6 0

3 years ago

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How many different license plates are possible if a license plate consists of 2 capital letters followed by 5 digits?

Alina [70]

The first letter can be any one of 26 letters. For each one . . .
The second letter can be any one of 26 letters. For each one . . .

The first digit can be any one of 10 digits. For each one . . .
The second digit can be any one of 10 digits. For each one . . .
The third digit can be any one of 10 digits. For each one . . .
The fourth digit can be any one of 10 digits. For each one . . .
The fifth digit can be any one of 10 digits.

The total number of possibilities is

(26 x 26 x 10 x 10 x 10 x 10 x 10) =

( 26² x 10⁵) = (676 x 100,000) = 67,600,000 .

6 0

4 years ago

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(x +y)^5 Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

3 years ago

(-5 + 5i) + (4 – 4i)?

babunello [35]

Answer:

The answer is:

Step-by-step explanation:

1. -30

2. 20

3 0

3 years ago

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A traffic officer is compiling information about the relationship between the hour or the day and the speed over the limit at wh

SpyIntel [72]

Answer:

The correlation between hour of the day and the speed over the limit at which the motorist is ticketed is weak positive correlation.

Step-by-step explanation:

The correlation coefficient is a statistical degree that computes the strength of the linear relationship amid the relative movements of the two variables (i.e. dependent and independent).It ranges from -1 to +1.

The types of correlation coefficient are:

+1 (-1) : Perfect positive (negative) correlation
0 to 0.30(-0.30) : Weak positive (negative) correlation
0.30(-0.30) to 0.70(-0.70) : Moderate positive (negative) correlation
0.70(-0.70) to 1 (-1) : Strong positive (negative) correlation

The correlation coefficient value between the hour of the day and the speed over the limit at which the motorist is ticketed is:

r = 0.12.

The value of r lies between:

0 < 0.12 < 0.30

Thus, the correlation between hour of the day and the speed over the limit at which the motorist is ticketed is weak positive correlation.

3 0

4 years ago