No need for alink i don't want a link i want a answer please i need hge help

PLEASE HELP HELP HELP

Jet001 [13]

Answer:

The answer is: 11

explanation: the powers get added 

If that is not correct it might be 30

i hope this was helpful :)

6 0

3 years ago

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experienced housepainter can paint an inside wall in under twenty minutes. An inexperienced housepainter can take up to one hour

vovikov84 [41]

1 hour
The exerience 3 walls in a hour and the other guy paint the otherone
Sorry i dont speek english

6 0

4 years ago

About A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows: A: The sum on the two

Flauer [41]

$p(A) = \frac{1}{2}$

$p(B) = \frac{1}{6}$

$p(C) = \frac{1}{6}$

$P(A | C)=\frac{1}{2}$

Solution:

The probability of an event is given as:

$\text { probability of an event }=\frac{\text { number of favorable outcomes }}{\text { total number of outcomes }}$

In throwing one die, the total number of outcomes = 6 { 1, 2, 3, 4, 5 , 6}

First let us calculate p(A):

The event is defined as: The sum on the two dice is even

Sum on two dice is even if and only if either both dice turn up odd or both even.

The odd outcomes in thowing a single die = 3 {1, 3, 5}

The even outcomes in throwing a single die = 3 {2, 4, 6}

The probability that both turn up odd is:

$\text { probability of both die turing up odd }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

Similarly, the probability that both turn up even is:

$\text { probability of both die turing up even }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

probability that the sum on two dice is even = probability that both turn up odd + probability that both turn up even

$\text { probability of sum on two dice is even }=\mathrm{p}(\mathrm{A})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Thus $p(A) = \frac{1}{2}$

Let us calculate p(B):

The event B is defined as: The sum on the two dice is at least 10

The total possible outcomes of two die is given as:

{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }

Since each individual die can turn up any of the numbers 1, 2, 3, 4, 5, 6 the event "sum of the two dice will be at least 10" is :

atleast 10 means that sum can be 10 or greater than 10

{(4,6), (6,4), (5,5), (5,6), (6,5), (6,6)}

Here favourable outcomes = 6

Total number of outcomes = 36

Hence, the probability that the sum of the two dice will be at least 10 is:

$\text { probability that the sum of the two dice will be at least } 10=\frac{6}{36}=\frac{1}{6}$

Thus $p(B) = \frac{1}{6}$

Let us calculate p(C):

The event C is defined as: The red die comes up 5

Favourable outcomes = {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}

$\text { probability of red die comes up } 5 \text { is the event }=\frac{6}{36}=\frac{1}{6}$

Thus $p(C) = \frac{1}{6}$

B) What is p(A l C)

$P(A | C)=\frac{p(A \cap C)}{P(C)}$

$\mathrm{A} \cap \mathrm{C}=\{(1,5),(3,5),(5,5)\}$

$p(A \cap C)=\frac{3}{36}=\frac{1}{12}$

$P(A | C)=\frac{p(A \cap C)}{P(C)}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}$

Thus $P(A | C)=\frac{1}{2}$

8 0

4 years ago

Simplify <img src="https://tex.z-dn.net/?f=%20%7B3%7D%5E%7B7y%7D%20%20%5Ctimes%20%20%7B81%7D%5E%7B8y%7D%20%5Ctimes%20%20%

Papessa [141]

Answer

$3^{7y} * 81^{8y} * 3^9 * 27^{-9y} = 3^{7y} * (3^4)^{8y} * 3^9 *(3^3)^{-9y}$

$=3^{7y} * (3)^{32y} * 3^9 *(3)^{-27y}\\\\=3^{7y +32y +9 -27y}\\\\=3^{12y+9}$

Tip :

$\frac{1}{27^{9y}} = 27^{-9y} = (3^3)^{-9y} = 3^{-27y}$

5 0

3 years ago

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What is the probability of drawing a diamond from a standard deck of cards on a second draw, given that a diamond was drawn on t

Charra [1.4K]

This problem is all about probability. It is the study of predicting the likelihood or chances of a certain event to happen out of all the possibilities. It is always expressed as a part of a whole. Therefore, the answer is either in fraction or in percentage.

A standard deck of card consists of 52 cards all in all. There are 13 diamond cards within the deck. So, the probability of getting a diamond card is 1/52. But we are given with a conditional probability. The first draw is sure to pick a diamond. So, the probability for this is 1 or 100%. But we should multiply this to the second scenario which is the 2nd draw. If you picked a card already, that means the total number of cards is 52 less 1. Also, because you already picked a diamond in the first draw, the diamond cards left in the deck is 13 less 1. Therefore, the probability of getting a diamond card in the second draw is

Probability = 1 × 12/51
Probability = 12/51

7 0

3 years ago