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3 years ago

11

Given that 3 feet equal 1 yard, how many yards equals 1 feet ?

1 answer:

never [62]3 years ago

6 0

Answer:

0.333333

Step-by-step explanation:

Hope this helps and have a great day!

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Use a graphing calculator to sketch the graph of the quadratic equation, and then give the coordinates for the x-

Umnica [9.8K]

Given:

The quadratic equation is

$y=6x^2-15x+6$

To find:

The x-intercepts of the given equation.

Solution:

We have,

$y=6x^2-15x+6$

Splitting the middle term, we get

$y=6x^2-12x-3x+6$

$y=6x(x-2)-3(x-2)$

$y=(x-2)(6x-3)$

For x-intercepts, y=0.

$(x-2)(6x-3)=0$

Using zero product property, we get

$x-2=0\text{ and }6x-3=0$

$x=2\text{ and }6x=3$

$x=2\text{ and }x=\dfrac{3}{6}$

$x=2\text{ and }x=0.5$

So, the x-intercepts are (0.5,0) and (2,0).

Therefore, the correct option is a.

5 0

2 years ago

Need help I'm lost and confused

Murljashka [212]

The answer is 89. it's really not that hard.juat follow the steps

8 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=2%20-%206%20%5Cdiv%206" id="TexFormula1" title="2 - 6 \div 6" alt="2 - 6 \div 6" align="absmid

Karolina [17]

2 - 6 ÷ 6 (Order of Operations - PEMDAS)

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7 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

can someone please help me solve these problems I did accept it on another paper I'm trying to see if I got them all correct I n

Hunter-Best [27]

3.
$3 \div 30 = 10$
sorry all I can help with at 10:15 at night

7 0

3 years ago