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3 years ago

What is the probability of pulling a spade or an ace from the deck?

Mathematics

2 answers:

nydimaria [60]3 years ago

7 0

for a spade to be pulled from a deck of cards is 13/52 chance.

for you to pull an ace from the deck is 4/52.

Yakvenalex [24]3 years ago

3 0

Well add the two together 13/52 + 4/52 = 17/52, but subtract out the ace of spades so you don’t count it twice, so 16/52 or 30.77%

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Which of the values shown are potential roots of f(x) = 3x3 – 13x2 – 3x + 45? Select all that apply.

galina1969 [7]

Answer:

All potential roots are 3,3 and $-\frac{5}{3}$ .

Step-by-step explanation:

Potential roots of the polynomial is all possible roots of f(x).

$f(x)=3x^3-13x^2-3x+45$

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

p=All the positive/negative factors of 45

q=All the positive/negative factors of 3

$p=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45$

$q=\pm 1,\pm 3$

All possible roots

$\frac{p}{q}=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45,\pm \frac{1}{3},\pm \frac{5}{3}$

Now we check each rational root and see which are possible roots for given function.

$f(1)= 3\times 1^3-13\times 1^2-3\times 1+45\Rightarrow 32\neq 0$

$f(-1)= 3\times (-1)^3-13\times (-1)^2-3\times (-1)+45\Rightarrow \neq 32$

$f(-3)= 3\times (-3)^3-13\times (-3)^2-3\times (-3)+45\Rightarrow \neq -144$

$f(3)= 3\times (3)^3-13\times (3)^2-3\times (3)+45\Rightarrow =0\\\\ \therefore x=3\text{ Potential roots of function}$

Similarly, we will check for all value of p/q and we get

$f(-5/3)=0$

Thus, All potential roots are 3,3 and $-\frac{5}{3}$ .

5 0

2 years ago

Read 2 more answers

Prove :<br>sin²θ + cos²θ = 1<br><br><br>thankyou ~

Gnom [1K]

Answer:

See below

Step-by-step explanation:

Here we need to prove that ,

$\sf\longrightarrow sin^2\theta + cos^2\theta = 1$

Imagine a right angled triangle with one of its acute angle as $\theta$ .

The side opposite to this angle will be perpendicular .
Also we know that ,

$\sf\longrightarrow sin\theta =\dfrac{p}{h} \\$

$\sf\longrightarrow cos\theta =\dfrac{b}{h}$

And by Pythagoras theorem ,

$\sf\longrightarrow h^2 = p^2+b^2 \dots (i)$

Where the symbols have their usual meaning.

Now , taking LHS ,

$\sf\longrightarrow sin^2\theta +cos^2\theta$

Substituting the respective values,

$\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\$

$\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\$

$\sf\longrightarrow \dfrac{p^2+b^2}{h^2}$

From equation (i) ,

$\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\$

$\sf\longrightarrow \bf 1 = RHS$

Since LHS = RHS ,

Hence Proved !

I hope this helps.

5 0

2 years ago

I'll give brainliest if u are right

emmainna [20.7K]

Answer:

0.3

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Savatey [412]

Answer:

Step-by-step explanation:

(X1, Y1) = (-3,-2)

(X2,Y2)=(4,8)

SLOPE= (Y2-Y1) / (X2-X1)

= (8-(-2)) / (4-(-3))

=(8+2) / (4+3)

= 10/7

Hope it was helpful

5 0

2 years ago

CAN SOMEbody please help me

tensa zangetsu [6.8K]

Https://www.calculatorsoup.com/calculators/geometry-solids/surfacearea.php
use this calculator it’s super helpful with lots of these kinds of questions!

4 0

2 years ago