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3 years ago

Is this right? If not let me know and explain how if u can. (No link <3)

Mathematics

1 answer:

vlabodo [156]3 years ago

8 0

Answer:

9/35

Step-by-step explanation:

There are a total of 15 bulbs to choose from randomly. First a new bulb is chosen. Chance is 9/15 because 9 out of 15 are new.

Then there are only 14 bulbs left. Then a used bulb is chosen. Chance is 6/14 since 6 out of remaining 14 are used.

Total chance for this to happen is 9/15 * 6/14 = 54/210 = 9/35.

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2^4. 2^5-(2^2)^2 Simplify the expression

11Alexandr11 [23.1K]

Answer:

Today: Monday, 12 October 2020

Hour: 17.31 WIB (in Indonesian)

_________________________

2⁴ × 2⁵ - (2²)²

= 2⁹ - 2⁴

= 512 - 16

= 496

7 0

3 years ago

Solve a = a+b+c+d/4 for c

kupik [55]

A=a+b+c+d/4
subtract a from both sides
0=b+c+d/4
Subtract b from both sides
-b=c+d/4
Finally subtract d/4 from both sides
c=-d/4-b

4 0

3 years ago

need help. solve for y. try factoring first if not possible or difficult use quadratic formula. y²-6y+6=0

lbvjy [14]

$y^2-6y+6=0\\ \\a=1 , \ b=-6, \ c=6 \\ \\\Delta =b^2-4ac = (-6)^2 -4\cdot1\cdot6 = 36-24=12\\ \\\sqrt{\Delta }= \sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3} \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{6-2\sqrt{3}}{2 }=\frac{2( 3- \sqrt{3})}{2}= 3- \sqrt{3}$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{6+2\sqrt{3}}{2 }=\frac{2( 3+ \sqrt{3})}{2}= 3+ \sqrt{3}$

3 0

3 years ago

Read 2 more answers

Soo-Jung used the subtraction property of equality to solve an equation for y. Which equation could be the one that Soo-Jung sol

Masteriza [31]

Answer:D y+4=12

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select

Blizzard [7]

Answer:

Probability is: $\frac{\textbf{13}}{\textbf{51}}$

Step-by-step explanation:

From a deck of 52 cards there are 26 black cards. (Spades and Clubs).

Also, there are 26 red cards. (Hearts and Diamonds).

First, we determine the probability of drawing a black card.

P(drawing a black card) = $\frac{number \hspace{1mm} of \hspace{1mm} black \hspace{1mm} cards}{total \hspace{1mm} number \hspace{1mm} of \hspace{1mm} cards}$ $= \frac{26}{52} = \frac{\textbf{1}}{\textbf{2}}$

Now, since we don't replace the drawn card, there are only 51 cards.

But the number of red cards is still 26,

∴ P(drawing a red card) = $\frac{number \hspace{1mm} of \hspace{1mm} red \hspace{1mm} cards}{total \hspace{1mm} number \hspace{1mm}of \hspace{1mm} cards}$ $= \frac{26}{51}$

Now, the probability of both black and red card = $\frac{1}{2} \times \frac{26}{51}$

$= \frac{\textbf{13}}{\textbf{51}}$

Hence, the answer.

5 0

3 years ago