All categories

2 years ago

A survey of 500 randomly selected people who commute to Manhattan for work showed that

Mathematics

1 answer:

mart [117]2 years ago

3 0

Answer:

187

Step-by-step explanation:

You might be interested in

Can someone please help me

djyliett [7]

Slope = (10 - 5)/(-9 + 15) = 5/6

y = mx + b where m = slope and b = y-intercept

b = y - mx

b = 5 - (5/6) (-15)

b = 5 +12.5

b = 17.5

Equation y = 5/6 x + 17.5

Y-intercept = 17.5

X-intercept when y = 0

5/6 x + 17.5 = 0

5/6 x = -17.5

x = (-17.5) (6/5)

x = -21

Answer

Y intercept (0, 17.5 )

X intercept (-21 , 0)

3 0

3 years ago

Which of the following expressions are equivalent to (x + 0) + y + 0

zzz [600]

answer c. none of the above

8 0

2 years ago

If JM = 5x – 8 and LM = 2x – 6, which expression represents JL?

muminat

Answer:

The expression represents JL is<u> 3x - 2</u>.

Step-by-step explanation:

Given:

JM = 5x – 8 and LM = 2x – 6.

Now, to find expression represents JL.

JM = JL + LM

<em>Subtracting both sides by LM we get:</em>

JM - LM = JL.

JL = JM - LM

Now, putting the expression to get JL:

$JL=5x-8-(2x-6)$

$JL=5x-8-2x+6$

$JL=3x-2.$

Therefore, the expression represents JL is 3x - 2.

8 0

3 years ago

X² – x² – 3x + 3<br> all rational zeros to a polynomial

Mrrafil [7]

Answer:

x = 1

Step-by-step explanation:

${x}^{2 } - {x}^{2} - 3x + 3 \\ - 3x + 3 \\ 3x = 3 \\ x = 1$

4 0

3 years ago

Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>

$x^ax^b = x^{(a+b)}$
$\dfrac{x^a}{x^b} = x^{(a-b)}$
$(a^a)^b =x^{(a\times b)}$
$(xy)^a = x^ay^a$

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

The solution of the expression is $\dfrac{4x^4}{y^6}$ .

<h2>Explanation</h2>

Given to us

$(16x^8y^{12})^{\frac{1}{2}}$

Solution

We know that 16 can be reduced to $2^4$ ,

$=(2^4x^8y^{12})^{\frac{1}{2}}$

Using identity $(xy)^a = x^ay^a$ ,

$=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}$

Using identity $(a^a)^b =x^{(a\times b)}$ ,

$=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})$

Solving further

$=2^2x^4y^{-6}$

Using identity $\dfrac{x^a}{x^b} = x^{(a-b)}$ ,

$=\dfrac{2^2x^4}{y^6}$

$=\dfrac{4x^4}{y^6}$

Hence, the solution of the expression is $\dfrac{4x^4}{y^6}$ .

Learn more about Exponentiation:

brainly.com/question/2193820

8 0

2 years ago