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Misha Larkins [42]
3 years ago
10

HELP worth 10 points!!

Mathematics
1 answer:
BaLLatris [955]3 years ago
4 0

Answer:

The answer is D 11/35

Step-by-step explanation:

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Wich is bigger<br><br> 1.320. 1.302<br><br> &lt;<br> &gt; <br> =
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1.320 >1.302 or 1.302<1.320
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3 years ago
Which of the following inequalities is graphed below?
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A : y is greater than or equal to 4x plus two
6 0
3 years ago
Matthew Casertano and Fox Chyatte make a series of bets. In each bet, Matthew sets the stake (the amount he wins or loses) at ha
Olin [163]

Answer:

219/256 = 0.855

Step-by-step explanation:

He starts with $256.

After the first bet, he either has $128 or $512.

After the second bet, he either has $64, $256, or $1024.

Repeating this pattern, the amount after n bets and r wins is:

A = 256 (½)ⁿ⁻ʳ (2)ʳ

And the number of ways A can be achieved is:

N = nCr

We're given that n = 8.

A = 256 (½)⁸⁻ʳ (2)ʳ

If r = 0, A = 1 and N = 1.

If r = 1, A = 4 and N = 8.

If r = 2, A = 16 and N = 28.

If r > 2, A > 50.  The sum of N for all r is 2⁸ = 256.

Of the 256 possible combinations, 1+8+28 = 37 result in less than $50, so 219 result in more than $50.

So the probability is 219/256 = 0.855.

5 0
3 years ago
Barron's reported that the average number of weeks an individual is unemployed is 18.5 weeks. Assume that for the population of
Ket [755]

Answer:

A) The sampling distribution for a sample size n=50 has a mean of 18.5 weeks and a standard deviation of 0.849.

B) P = 0.7616

C) P = 0.4441

Step-by-step explanation:

We assume that for the population of all unemployed individuals the population mean length of unemployment is 18.5 weeks and that the population standard deviation is 6 weeks.

A) We take a sample of size n=50.

The mean of the sampling distribution is equal to the population mean:

\mu_s=\mu=18.5

The standard deviation of the sampling distribution is:

\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{50}}=0.849

B) We have to calculate the probability that the sampling distribution gives a value between one week from the mean. That is between 17.5 and 19.5 weeks.

We can calculate this with the z-scores:

z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{17.5-18.5}{6/\sqrt{50}}=\dfrac{-1}{0.8485}=-1.179\\\\\\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{19.5-18.5}{6/\sqrt{50}}=\dfrac{1}{0.8485}=1.179

The probability it then:

P(|X_s-\mu_s|

C) For half a week (between 18 and 19 weeks), we recalculate the z-scores and the probabilities:

z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{18-18.5}{6/\sqrt{50}}=\dfrac{-0.5}{0.8485}=-0.589

P(|X_s-\mu_s|

5 0
3 years ago
HOW DO YOU DO THIS? PLEASE SHOW WORK ASAP!
MrMuchimi
2. m<-1.5
3. 11≥x
4. x≥1
5. x≥2

5 0
3 years ago
Read 2 more answers
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