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MA_775_DIABLO [31]
3 years ago
7

A regression by a sample of 10 observations gives: Score=49.5+1.96 Study Hours. The sample mean of study hour is 10.5 and sum of

the squared deviations of this independent variable is 264.5. Now, what is the leverage statistic for study hour=14?
Mathematics
1 answer:
amm18123 years ago
4 0

Answer:

0.1463

Step-by-step explanation:

Number of observations = 10

Sample mean = 10.5

Sum of standard deviation = 264.5

X = 14

We are to calculate The leverage statistic for study hour 14 using the data above

= 1/10 + (14-10.5)²/264.5

= 1/10 + 3.5²/264.5

= 0.1 +12.25/264.5

= 0.1+0.04631

= 0.1463 is the leverage statistic for the study

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umka2103 [35]
She will need 8 pints

2 cups in 1 pint so 

16/2=8 
5 0
3 years ago
Please help me fast i am giving 14 points for 5 questions. (Geometry) Thank you so much!
algol13
1 ) Area of the rectangle:
A = L x W
L = √(2² + 2²) = √8 = 2√2
W = √(6² + 6²) = √72 = 6√2
A = 2√2  x  6√2 = 24 units²
2 ) Area of a triangle:
RQ = 2 + 4 = 6 units
h = 4 units
A = ( 6 * 4 ) / 2 = 12 units²
3 ) The perimeter of Δ ABC:
AB = √(3² + 4²) = √25 = 5 units
BC = √(1² + 1²) = √2 = 1.4 units
AC = √(3² + 4²) = √25 = 5 units
P = 5 + 1.4 + 5 = 11.4 units
4 ) Area of the figure ( approx.):
A ≈ ( 8 * 8) - 6.25 - 8 - 2.5 ≈ 47.25
Answer: C ) 50 ft²
5 ) Area under the curve:
A ≈ 0.5 * 3 + 0.5 * 3.5 + 0.5 * 4 + 0.5 * 4.5 + 0.5 * 5 + 0.5 * 4.5 + 0.5 * 4 +
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6 0
3 years ago
g An urn contains 150 white balls and 50 black balls. Four balls are drawn at random one at a time. Determine the probability th
xxTIMURxx [149]

Answer:

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

Step-by-step explanation:

For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Sampling with replacement:

I consider a success choosing a black ball, so p = \frac{50}{150+50} = \frac{50}{200} = 0.25

We want 2 black balls and 2 white, 2 + 2 = 4, so n = 4, and we want P(X = 2).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(0.25)^{2}.(0.75)^{2} = 0.2109

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Sampling without replacement:

150 + 50 = 200 total balls, so N = 200

Sample of 4, so n = 4

50 are black, so k = 50

We want P(X = 2).

P(X = 2) = h(2,200,4,50) = \frac{C_{50,2}*C_{150,2}}{C_{200,4}} = 0.2116

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

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3 years ago
Find x <br><br>Plz help me!!!!
lbvjy [14]
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3 years ago
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Lady bird [3.3K]

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- 15° - 15°

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I am joyous to assist you anytime.

3 0
3 years ago
Read 2 more answers
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