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Alisiya [41]
3 years ago
14

Can someone help me please ?

Mathematics
1 answer:
pantera1 [17]3 years ago
7 0

Answer:

-5/-3

Step-by-step explanation:

first y intercept=

-5/-3

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Zepler [3.9K]

Answer:

Hope this helps

Step-by-step explanation:

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QUESTION 1
podryga [215]

Answer: The team won 47 games.

Step-by-step explanation: The total games played was 84, and that means all games added together, but victories and losses. If the games they lost is called X, and the team won 10 more games than it lost, then it means the games they won is X + 10.

To find the value of wins and losses, we can now express both properly as;

X + (X +10) = 84

2X + 10 = 84

Subtract 10 from both sides of the equation

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If the number of games won is X + 10, then games won is derived as

Wins = X + 10

Wins = 37 + 10

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The team won 47 games

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3 years ago
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Describing the Vertical Line Test<br> Explain what the vertical line test is and how it is used.
Radda [10]

Answer:

Vertical line test is used to identify that the given graph is function or not

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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.

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2 years ago
Which expression is equivalent to 7m + 5 + 6 + 3m
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Answer:21

Step-by-step explanation:

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