To the nearest tenth, the P( -2.1 < z

Which lists these numbers in order from least to greatest? –14.6, 159, 1.07, –1.295, 24.6

Monica [59]

c is the answer because negative numbers have to go first

4 0

2 years ago

What is the equation for the line of reflection that maps the trapezoid onto itself? Enter your answer in the box______

nadezda [96]

Y = -1

It is an isosceles trapezoid, which is just as well, otherwise it could not reflect onto itself. The line of reflection is the same as the mirror line which is the same as the axis of symmetry.

It runs horizontally exactly through the middle of the shape.

All horizontal lines have the equation y = a value

In this case the line is y=−1

6 0

3 years ago

In a survey, 150 people were asked if they smoke and whether or not they exercise regularly (more than twice a week). The result

s2008m [1.1K]

(a) 28%

(b) 24%

When calculating the total amount of people it come out to 150.

According to the table 42 people smoke. 42/150 is 28%

According to the table 36 people exercise regularly. 36/150 is 24%

7 0

2 years ago

(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

2 years ago

The average of 4 numbers is 5. What is their sum? explain you answer

maxonik [38]

To find an average you add all numbers together then divide by the amount of numbers that you added. So if there are 4 numbers you would divide by 4. By going backwards you can find the answer. Multiply 5 by 4 and you get 20, which is the answer.

8 0

3 years ago

To the nearest tenth, the P( -2.1 < z < 0) is: