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ruslelena [56]
3 years ago
13

For which value of bb is x2+bx−60x2+bx-60 factorable over the set of integers

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
3 0
Hsjakakskskkskskd. because ypu never know athaysne s
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A soccer game lasts 90 minutes without running into overtime. The coach
KatRina [158]

Answer: 7 Switches will be made.

Step-by-step explanation:

\frac{90}{12.5}= number of switches

7.2= number of switches

  • however you cannot have a fraction of a switch so you must round it to a whole number, which in this case is 7.

5 0
2 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
Need help ASAP!!!!!!!!!!!!!!!!!!!!!!!!!
il63 [147K]

Answer:

d

Step-by-step explanation:

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5 0
3 years ago
List the sections of a business plan
Salsk061 [2.6K]

Answer:

Executive Summary. ...

Company Description. ...

Products and Services. ...

Market analysis: ...

Strategy and Implementation: ...

Organization and Management Team: ...

Financial plan and projections:

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Good evening! Can someone please answer this, ill give you brainliest and your earning 50 points. Would be very appreciated.
Tanzania [10]

Tyler concludes that 5x² will always have a larger output for the same value of x.

<u>Look at the graph below and the table given</u>

Take a random value: x = 0

  • When 5x², y = 0
  • When 2^x, y = 1

Here, 1 > 0, making 2^x > 5x²

Hence, 2^x is greater than 5x² at this point. making Tyler's point not applicable.

Disagree with Tyler's point.

7 0
2 years ago
Read 2 more answers
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