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babunello [35]
3 years ago
6

X − 5 = 17 A) x = 22 B) x = 12 C) x = 3.4 D) x = -12

Mathematics
1 answer:
Maksim231197 [3]3 years ago
8 0
A is the correct answer coz if you substitute 22 in the place of X-5 the answer will be 17
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A sample of 18 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is n
ki77a [65]

Answer:

a)

i.  x=- - - - - -

ii.  o=- - - - - -

iii.  Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.  

c) The statistics variable X  is a measure on a sample that is used as an estimate of the population mean.X  is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters  because the population standard deviation is known.

N(X, o/\sqrt{n})

So, the distribution is  

N(2, 0.12/\sqrt{16})

e)

i. The 90% confidence interval for the population mean weight of the candies is  1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is  1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part  is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

Step-by-step explanation:

a)

i.  x=- - - - - -

ii.  o=- - - - - -

iii.  Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.  

c) The statistics variable X  is a measure on a sample that is used as an estimate of the population mean.X  is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters  because the population standard deviation is known.

N(X, o/\sqrt{n})

So, the distribution is  

N(2, 0.12/\sqrt{16})

e)

i. The 90% confidence interval for the population mean weight of the candies is  1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is  1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part  is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

5 0
2 years ago
What is the range if this function ?
Pavel [41]

Answer:

-4 to 0 (both -4 and 0 are included in the range because of the dots)

3 0
2 years ago
Please Help Me!!!!!!!​
Ghella [55]

Answer:

D

Step-by-step explanation:

(6, 24)  this ordered pair has an x-value of 6 and a y-value of 24

The only equation, if you substitute those values into it, which works is D.

24 = 4(6)

24 = 24

6 0
3 years ago
Need Answer to these questions. pls help
vredina [299]

Answer:

7.2 million dollars

Step-by-step explanation:

The domain of the function the complete set of possible values of the independent variable. In this case the domain includes all real numbers except 189.

From;

T(p) = 0.50 (p - 189)

When p = 203.4 million dollars

T(p) = 0.50 (203.4 - 189)

T(p) = 7.2 million dollars

8 0
2 years ago
Https://forms.gle/NYohXkMYJGTVQDkf6
weeeeeb [17]

Answer:

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Step-by-step explanation:

never click links unless you know EXACTLY whats on the other side folks

stay safe in the cyber-verse :D

3 0
2 years ago
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