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Dovator [93]
3 years ago
6

Math Help

Mathematics
1 answer:
-Dominant- [34]3 years ago
4 0
X=1, because they both have the same height (y), so they'll cross each other!
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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
3 years ago
Will give brainliest and 5 stars
BaLLatris [955]
First, find what percentage of students had 3 or more by adding up your known percents:

45% + 23 % + 21% + x% = 100%
x = 11%

Since you're given that 96 students had 2 or more, you add up the percentages of 2 and 3 or more:

11 + 21 = 32%

Now set up a proportion that relates it to the whole:

\frac{96 students}{32 percent} = \frac{x students}{100 percent}

This will allow you to find the total number of students at the school.

Cross multiplying and solving for x results in 300 total students.

Question 1:
45% had one or more absences. 45% of 300 students is 135 students.

Question 2:
As we found before, 11% of students had three or more. 11% of 300 is 33 students.
5 0
4 years ago
The Pennington and Williams family each went on road trips this summer. Their mileage is shown in the graph
hoa [83]
Take a picture of the graph
8 0
3 years ago
The high school took two field trips to the Newport Aquarium many many years ago. They used school buses and vans to transport a
il63 [147K]

Answer:

bus = 43

van 12

Step-by-step explanation:

This can be solved using simultaneous equations

Let v represent the number of students that a van carries

Let b represent the number of students that a bus carries

the following equations can be derived from the question

3v + 2b = 122 eqn 1

5v + 3b = 189  eqn 2

Multiply eqn 1 by 5 and eqn 2 by 3

15v + 10b = 610  eqn 3

15v + 9b = 567 eqn 4

Subtract equation 4 from 3

b = 43

Substitute for b in equation 1

3v + 2(43) = 122

solve for v

v = 12

5 0
3 years ago
Find the slope plz help ASAP!!
vlada-n [284]
Up 3 left 4
Slope= -3/4
3 0
3 years ago
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