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3 years ago

12

This spinner has 6 equal sections. What are the odds in simplest form against spinning blue or yellow?

2 answers:

astra-53 [7]3 years ago

5 0

Answer:

so the answer is A,B

Rolling two six-sided dice: Each die has 6 equally likely outcomes, so the sample space is 6 • 6 or 36 equally likely outcomes. Flipping three coins: Each coin has 2 equally likely outcomes, so the sample space is 2 • 2 • 2 or 8 equally likely outcomes.

Step-by-step explanation:

MariettaO [177]3 years ago

3 0

B) Th chance of getting yellow and blue would be 2/6 but not getting it would be 4/6 and 4/6 in its simplest form is 2/3

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Solve x2 – 12x + 36 = 90 for x.

Hitman42 [59]

A: x= 6+3 squareroot 10 ... Its letter A

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3 years ago

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Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

7 0

3 years ago

In order for cos 53° = sin A, what must be the measure of _A?<br> =<br> 53°<br> 37°<br> 127°<br> 90°

Ray Of Light [21]

Answer:

the answer is 37..............

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2 years ago

The sum of the numbers as a product of their GCF is ? The numbers are18+48

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

18=2×3×3

48=2×2×2×2×3

G.C.F.=2×3=6

18+48=66

6×11=66

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3 years ago

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zhuklara [117]

1/12

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3 years ago