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suter [353]
2 years ago
11

....................... Plz​

Mathematics
2 answers:
kramer2 years ago
7 0

Answer:

1. x=77/2

2. m=2

Step-by-step explanation:

vesna_86 [32]2 years ago
6 0

Answer:

2/5 (x-1)=15 (x=7)

7 (1+3m)=49 (m=2)

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A snack bar offers chili and or cheese on its hot dogs for an additional fee. the owner kept track of the hot dog orders over th
kozerog [31]

30% of his customers ordered chili with no cheese

5 0
3 years ago
Please please please please help
navik [9.2K]

Answer:

14

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Suppose that an internal report submitted to the managers at a bank in Boston showed that with 95 percent confidence, the propor
Dmitry [639]

Answer:

The sample size used to compute the 95% confidence interval is 1066.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The 95% confidence interval for proportion of the bank's customers who also have accounts at one or more other banks is (0.45, 0.51).

To compute the sample size used we first need to compute the sample proportion value.

The value of sample proportion is:

\hat p=\frac{Upper\ limit+Lower\ limit}{2}=\frac{0.45+0.51}{2}=0.48

Now compute the value of margin of error as follows:

MOE=\frac{Upper\ limit-Lower\ limit}{2}=\frac{0.51-0.45}{2}=0.03

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the value of sample size as follows:

MOE=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\0.03=1.96\times \sqrt{\frac{0.48(1-0.48)}{n}}\\(\frac{0.03}{1.96})^{2}=\frac{0.48(1-0.48)}{n}\\n=1065.404\\n\approx1066

Thus, the sample size used to compute the 95% confidence interval is 1066.

4 0
2 years ago
You play the following game against your friend. You have 2 urns and 4 balls One of the balls is black and the other 3 are white
Rom4ik [11]

Answer:

Part a: <em>The case in such a way that the chances are minimized so the case is where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b: <em>The case in such a way that the chances are maximized so the case  where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c: <em>The minimum and maximum probabilities of winning  for n number of balls are  such that </em>

  • <em>when all the n balls are placed in one of the urns the probability of the winning will be least as 1/2n</em>
  • <em>when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, as 0.5</em>

Step-by-step explanation:

Let us suppose there are two urns A and A'. The event of selecting a urn is given as A thus the probability of this is given as

P(A)=P(A')=0.5

Now the probability of finding the black ball is given as

P(B)=P(B∩A)+P(P(B∩A')

P(B)=(P(B|A)P(A))+(P(B|A')P(A'))

Now there can be four cases as follows

Case 1: When all the four balls are in urn A and no ball is in urn A'

so

P(B|A)=0.25 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.25*0.5)+(0*0.5)

P(B)=0.125;

Case 2: When the black ball is in urn A and 3 white balls are in urn A'

so

P(B|A)=1.0 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1*0.5)+(0*0.5)

P(B)=0.5;

Case 3: When there is 1 black ball  and 1 white ball in urn A and 2 white balls are in urn A'

so

P(B|A)=0.5 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.5*0.5)+(0*0.5)

P(B)=0.25;

Case 4: When there is 1 black ball  and 2 white balls in urn A and 1 white ball are in urn A'

so

P(B|A)=0.33 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.33*0.5)+(0*0.5)

P(B)=0.165;

Part a:

<em>As it says the case in such a way that the chances are minimized so the case is case 1 where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b:

<em>As it says the case in such a way that the chances are maximized so the case is case 2 where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c:

The minimum and maximum probabilities of winning  for n number of balls are  such that

  • when all the n balls are placed in one of the urns the probability of the winning will be least given as

P(B|A)=1/n and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/n*1/2)+(0*0.5)

P(B)=1/2n;

  • when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, equal to calculated above and is given as

P(B|A)=1/1 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/1*1/2)+(0*0.5)

P(B)=0.5;

5 0
3 years ago
In the function f(x)=6x^2-4, what is the value of f(-2)?
BartSMP [9]

In order to find the value of f(-2), you can plug -2 wherever x is.

f(x) = 6x^2 - 4

f(-2) = 6(-2)^2 - 4

f(-2) = 6(4) - 4

f(-2) = 24 - 4

f(-2) = 20

Thus, the value of f(-2) is 20. Hope this helps! If you have any other questions related to this, let me know.

4 0
3 years ago
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