A siberian Tiger can grow 350 centimeters long, excluding the tail. How long is this in feet?

The heights of 200 adults were recorded and divided into two categories. Which two way frequency table correctly shows the margi

Leto [7]

Total men is 12 + 86 = 98

If 200 people were surveyed, that means the number of women should be 200 - 98 = 102.

Table B is the only one with 102 total women.

8 0

2 years ago

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Help me please thank you

Lapatulllka [165]

All of them except TR and PH

3 0

2 years ago

What is the equation for the linear model in the scatter plot obtained by choosing the two points to the closest line?

sergiy2304 [10]

(4,12) (20,36)
slope = (36 - 12)/(20-4) = 24 / 16 = 3/2 = 1.5

y intercept (0,6) b = 6
equation
y = 1.5x + 6

answer
b. y = 1.5x + 6

4 0

2 years ago

Raphael graphed the functions g(x)=x+2 and f(x)=x−1. How many units below the y-intercept of g(x) is the y-intercept of f(x)? A

jasenka [17]

Answer:

3 units below the y-intercept of g(x) is the y-intercept of f(x)

Step-by-step explanation:

$g(x)=x+2$ and $f(x)=x-1$

In f(x)= mx+b the y intercept is b

In $g(x)=x+2$ , the y intercept value is 2

In $f(x)=x-1$ , the y intercept value is -1

the difference in y intercept is +2-(-1)=+3

3 units below the y-intercept of g(x) is the y-intercept of f(x)

4 0

2 years ago

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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago