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guajiro [1.7K]
3 years ago
10

MARK AS BRAINLIEST!!!

Mathematics
1 answer:
liq [111]3 years ago
6 0

Answer:

Part A

it supposed to be x⩾-1/3 since you divided by negative

Step-by-step explanation:

Part A

4-6x⩾-15x+1

step 1: add -6x to both sides

4-6x+6x⩾-15x+6x+1

=4⩾-9x+1

step 2: subtract 1 from both sides

4-1⩾-9x+1-1

3⩾-9x

step 3: do reflexive property and divide

3⩾-9x

= <u>-9x⩽3</u>

-9

x⩽-1/3

and since you divided by negative, the sign must change. so it'll be x⩾-1/3

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2 + 3(3x - 6) = 5(x - 3) + 15 I'm in Algebra 1 and I still can't seem to get this problem, I use Slader to find how to do this b
Wewaii [24]

Answer:

x = 4

Step-by-step explanation:

2 + 3(3x - 6) = 5(x - 3) + 15

Expand the parenthesis:

2 + 9x - 18 = 5x - 15 + 15

Simplify:

9x - 16 = 5x

Subtract 5x from both sides:

4x - 16 = 0

Add 16 to both sides:

4x = 16

Divide both sides by 4:

x = 4

7 0
3 years ago
tasha has 24 new pieces of art. 1/6 are photo's and 2/3 are paintings. how many more paintings are there than photos?
ella [17]
0.5 x 24 = 12
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3 years ago
Expand and combine like terms. (4b²+3)(4b²-3)​
Ad libitum [116K]

Answer:

16b4=9

Step-by-step explanation:

(4b²+3) (4b²-3)

16b4-12b²+12b²-9

16b(raised to power 4)-9

16b4-9

collect like terms

16b4=9

8 0
3 years ago
There are two games involving flipping a coin. In the first game you win a prize if you can throw between 45% and 55% heads. In
Nina [5.8K]

Answer:

d) 300 times for the first game and 30 times for the second

Step-by-step explanation:

We start by noting that the coin is fair and the flip of a coin has a probability of 0.5 of getting heads.

As the coin is flipped more than one time and calculated the proportion, we have to use the <em>sampling distribution of the sampling proportions</em>.

The mean and standard deviation of this sampling distribution is:

\mu_p=p\\\\ \sigma_p=\sqrt{\dfrac{p(1-p)}{N}}

We will perform an analyisis for the first game, where we win the game if the proportion is between 45% and 55%.

The probability of getting a proportion within this interval can be calculated as:

P(0.45

referring the z values to the z-score of the standard normal distirbution.

We can calculate this values of z as:

z_H=\dfrac{p_H-\mu_p}{\sigma_p}=\dfrac{(p_H-p)}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_H-p)>0\\\\\\z_L=\dfrac{p_L-\mu_p}{\sigma_p}=\dfrac{p_L-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_L-p)

If we take into account the z values, we notice that the interval increases with the number of trials, and so does the probability of getting a value within this interval.

With this information, our chances of winning increase with the number of trials. We prefer for this game the option of 300 games.

For the second game, we win if we get a proportion over 80%.

The probability of winning is:

P(p>0.8)=P(z>z^*)

The z value is calculated as before:

z^*=\dfrac{p^*-\mu_p}{\sigma_p}=\dfrac{p^*-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p^*-p)>0

As (p*-p)=0.8-0.5=0.3>0, the value z* increase with the number of trials (N).

If our chances of winnings depend on P(z>z*), they become lower as z* increases.

Then, we can conclude that our chances of winning decrease with the increase of the number of trials.

We prefer the option of 30 trials for this game.

8 0
3 years ago
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Ede4ka [16]

Answer:

(5, 4 )

Step-by-step explanation:

Given the 2 equations

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y = 4

solution is (5, 4 )

7 0
3 years ago
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