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3 years ago

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I have 2 bags of counters. The first bag contains 2 red counters and 1 blue counter. The second bag contains 1 red counter, 1 bl

ue counter and 2 yellow counters. I take a counter at random from both bags, what is the probability that two counters will be the same colour?

1 answer:

ASHA 777 [7]3 years ago

5 0

3/12, i used a tree diagram to work it out

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One teacher wants to give each student 1 1/5 slices of pizza. If the teacher has 12 slices of pizza, then how many students will

yKpoI14uk [10]

Answer:

10

Step-by-step explanation:

You just have to convert the fraction into numerical form.

1 and 1/5

change it to

1 and 2/10

so 1.2

You then divide 12 by 1.2 which gives you 10.

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3 years ago

Can you tell me what 6.488 rounds to

Kay [80]

Rounding<span> 6.488 up to the tenth place would be 6.5</span>

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3 years ago

? Which one? 0wo nhfhffbgsdh

BabaBlast [244]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet

frosja888 [35]

There seems to be one character missing. But I gather that <em>f(x)</em> needs to satisfy

• $x^3$ divides $f(x)$

• $(x-1)^3$ divides $f(x)^2$

I'll also assume <em>f(x)</em> is monic, meaning the coefficient of the leading term is 1, or

$f(x) = x^5 + \cdots$

Since $x^3$ divides $f(x)$ , and

$f(x) = x^3 p(x)$

where $p(x)$ is degree-2, and we can write it as

$f(x)=x^3 (x^2+ax+b)$

Now, we have

$f(x)^2 = \left(x^3p(x)\right)^2 = x^6 p(x)^2$

so if $(x - 1)^3$ divides $f(x)^2$ , then $p(x)$ is degree-2, so $p(x)^2$ is degree-4, and we can write

$p(x)^2 = (x-1)^3 q(x)$

where $q(x)$ is degree-1.

Expanding the left side gives

$p(x)^2 = x^4 + 2ax^3 + (a^2+2b)x^2 + 2abx + b^2$

and dividing by $(x-1)^3$ leaves no remainder. If we actually compute the quotient, we wind up with

$\dfrac{p(x)^2}{(x-1)^3} = \underbrace{x + 2a + 3}_{q(x)} + \dfrac{(a^2+6a+2b+6)x^2 + (2ab-6a-8)x +2a+b^2+3}{(x-1)^3}$

If the remainder is supposed to be zero, then

$\begin{cases}a^2+6a+2b+6 = 0 \\ 2ab-6a-8 = 0 \\ 2a+b^2+3 = 0\end{cases}$

Adding these equations together and grouping terms, we get

$(a^2+2ab+b^2) + (2a+2b) + (6-8+3) = 0 \\\\ (a+b)^2 + 2(a+b) + 1 = 0 \\\\ (a+b+1)^2 = 0 \implies a+b = -1$

Then $b=-1-a$ , and you can solve for <em>a</em> and <em>b</em> by substituting this into any of the three equations above. For instance,

$2a+(-1-a)^2 + 3 = 0 \\\\ a^2 + 4a + 4 = 0 \\\\ (a+2)^2 = 0 \implies a=-2 \implies b=1$

So, we end up with

$p(x) = x^2 - 2x + 1 \\\\ \implies f(x) = x^3 (x^2 - 2x + 1) = \boxed{x^5-2x^4+x^3}$

6 0

2 years ago

PLEASE HELP<br><br><br><br>this is a formula for a sequence, f(x) = 8 - 11x, what is f(8)?

Elenna [48]

Answer:

-80

Step-by-step explanation:

plug in 8 to the equation

8 - 11 (8)

multiply 8 and 11 first: 8 - 88

add 8 to -88: -80

7 0

3 years ago