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Jet001 [13]
3 years ago
5

In a children’s book, the mean word length is 3.6 letters with a standard deviation of 2.1 letters. In a novel aimed at teenager

s, the mean word length is 4.4 letters with a standard deviation of 2.4 letters. Both distributions of word length are unimodal and skewed to the right. Independent random samples of 40 words are selected from each book. Let xC represent the sample mean word length in the children’s book and let xT represent the sample mean word length in the teen novel.
Find the mean of the sampling distribution of xC-xT
Calculate and interpret the standard deviation of the sampling distribution. Verify that the 10% condition is met.
Justify that the shape of the sampling distribution is approximately Normal.
What is the probability that the sample mean word length is greater in the sample from the children’s book than in the sample from the teen novel?
Mathematics
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

Find the mean of the sampling distribution of xC-xT

Calculate and interpret the standard deviation of the sampling distribution. Verify that the 10% condition is met.

Justify that the shape of the sampling distribution

Step-by-step explanation:

2.4 letters. Both distributions of word length are unimodal and skewed to the right. Independent random samples of 40 words

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2 years ago
The sum of two numbers is 40. If the larger is divided by the smaller, the quotient is 5 and the remainder is 4. Find the number
solmaris [256]

Answer:

numbers are 33 and 7

Step-by-step explanation:

x+y=40

4y+5+y=40

5y+5=40

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6 0
3 years ago
Item 16 An indoor water park has two giant buckets that slowly fill with 1000 gallons of water before dumping it on the people b
nignag [31]

Answer:

  • 3:16 P.M.
  • 5:22 P.M.

Step-by-step explanation:

The two dump intervals have a greatest common factor (GCF) of 3, so their least common multiple (LCM) is ...

  (18)(21)/3 = 126 . . . . minutes

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4 0
3 years ago
How do i do 3 part a ?
WINSTONCH [101]


In general the binomial expansion is


(a+b)^n = {n \choose 0} a^0 b^n + {n \choose 1} a^1 b^{n-1} + {n \choose 2} a^2 b^{n-2} + ... + {n \choose n} a^n b^0


So in our case, because we want ascending powers of x we'll write,


(-3x + 1)^{11} =  {11 \choose 0} (-3x)^0 1^{11} + {11 \choose 1} (-3x)^{1} 1^{10} + {11 \choose 2} (-3x)^{2} 1^9  + {11 \choose 3 } (-3x)^3 1^8 + ...


We need to calculate the binomial coefficients:


{11 \choose 0}  = 1


{11 \choose 1}  = 11


{11 \choose 2}  = \dfrac{11 \times 10}{2} = 55


{11 \choose 3}  = \dfrac{11 \times 10 \times 9}{3 \times 2} = 165


(-3x+1)^{11} =  1 (-3x)^0 1^{11}  + 11(-3x)^{1} 1^{10}  + 55 (-3x)^2 1^{9}  + 165 (-3x)^3 1^8 + ...


(1-3x)^{11} =  1 -33 x + 495 3x^2 - 4455 x^3+ ...



6 0
3 years ago
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