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tekilochka [14]
3 years ago
11

A fair attended by 755,082 people altogether.What is the number rounded to the nearest ten thousand

Mathematics
1 answer:
Fudgin [204]3 years ago
7 0

Answer:

760,000


Step-by-step explanation:


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3 years ago
1.
GREYUIT [131]

Answer:

Step-by-step explanation:

1.

James goes to an arcade.

He has one go on the Teddy Grabber.

He has one go on the Penny Drop.

The probability that he wins on the Teddy Grabber is 0.2.

The probability that he wins on the Penny Drop is 0.3.

(a) Complete the tree diagram.

Teddy Grabber

Penny Drop

Win

0.3

Win

0.2

Lose

Win

Lose

Lose

(2)

7 0
3 years ago
How much is 3×4+9÷6+44×77×777×00+9?​
jarptica [38.1K]

Answer:3*9+9/6+44*77*777*00+9, so using order of applications, it simplifies to 27+1.5+9 after multiplying and dividing, and the final answer is 37.5

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Translate the verbal phrase to an equation, Two eighths of a number minus 4 is 2.
erastovalidia [21]

Answer:

2/8(x) - 4 = 2

<em>Hope this helps! </em>

8 0
3 years ago
NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!! THIS IS NOT A TEST OR AN ASSESSMENT!! Please help me with these math questions
Art [367]

Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector \vec{v} in some vector space \mathbb R^n we have

\vec{v} = \langle a,b\rangle

This is the component form. I don't like that way. It is probably used in high school, but

\vec{v} =  \begin{pmatrix} a\\ b\\ \end{pmatrix}

is preferable because the inner product on \mathbb R^n is defined to be

$\langle a,b\rangle := \sum_{i = 1}^n a_i b_i$

You can also write it using linear form such as \vec{v} = 2i+2j

4.

For this question, I think you meant

vectors

\vec{u_1} = (-8, 12)

\vec{u_2}  = (13, 15)

Once

\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}

Considering that the dot product is

\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76

and the norm of \vec{u_1} is ||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}

and the norm of \vec{u_2} is ||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}

Thus,

\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}

\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)

3 0
2 years ago
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