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3 years ago

I have no idea what 6/x = 8/19 is help bruv

Mathematics

1 answer:

bazaltina [42]3 years ago

8 0

Answer: X=17

Step-by-step explanation: (sorry if im wrong) so basicly X would be 17 because 6+2=8 so X has to +2 to get to 19 making X=17

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A small city has three automobile dealerships: a GM dealer selling Chevrolets and Buicks; a Ford dealer selling Fords and Lincol

9966 [12]

Answer:

Event A = { Chevrolet , Buick }

Event B = { Ford , Lincoln }

Event C = { Toyota }

Step-by-step explanation:

- Mutually exclusive events are such that their probability of coming true simultaneously is zero. If we consider set notations we could say.

P (A & B) = P (B & C) = P (A & C) = 0

- In our case these events A,B, and C can be defined as:

Answer:

Event A = { Chevrolet , Buick }

Event B = { Ford , Lincoln }

Event C = { Toyota }

4 0

3 years ago

Factor completely 4x2 − 32.

Korvikt [17]

Answer: 4(x + 4)(x - 4)

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Please help

iren2701 [21]

For question 1:
1% = 120 ÷ 100 = 1.2
15% = 1.2 × 15 = 18
So, 15% of 120 is 18

For question 2:
68.4 ÷ 72 = 0.95
0.95 × 100 = 95%
So, the percentage is 95%

For question 3,
1% = 12.5 ÷ 8 = 1.5625
100% = 1.5625 × 100 = 156.25
So, 8% of 12.5 is 156.25.

For question 4,
1% = 48 ÷ 40 = 1.2
100% = 1.2 × 100 = 120
So, 40% of 48 is 120.

For question 5,
1% = 195 ÷ 100 = 1.95
62.5% = 1.95 × 62.5 = 121.875
So, 62.5% of 195 is 121.875

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

For the data set 3,3,9,12,13, the mean is 8.What is the mean absolute deviation?

aleksandrvk [35]

The answer is C: The MAD is 4. Hope that helps

8 0

3 years ago

Read 2 more answers