Answer: 
<u>Step-by-step explanation:</u>
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Complete question :
Wright et al. [A-2] used the 1999-2000 National Health and Nutrition Examination Survey NHANES) to estimate dietary intake of 10 key nutrients. One of those nutrients was calcium in all adults 60 years or older a mean daily calcium intake of 721 mg with a standard deviation of 454. Usin these values for the mean and standard deviation for the U.S. population, find the probability that a randonm sample of size 50 will have a mean: (mg). They found a) Greater than 800 mg b) Less than 700 mg. c) Between 700 and 850 mg.
Answer:
0.10935
0.3718
0.9778
0.606
Step-by-step explanation:
μ = 721 ; σ = 454 ; n = 50
P(x > 800)
Zscore = (x - μ) / σ/sqrt(n)
P(x > 800) = (800 - 721) ÷ 454/sqrt(50)
P(x > 800) = 79 / 64.205295
P(x > 800) = 1.23
P(Z > 1.23) = 0.10935
2.)
Less than 700
P(x < 700) = (700 - 721) ÷ 454/sqrt(50)
P(x < 700) = - 21/ 64.205295
P(x < 700) = - 0.327
P(Z < - 0.327) = 0.3718
Between 700 and 850
P(x < 850) = (850 - 721) ÷ 454/sqrt(50)
P(x < 850) = 129/ 64.205295
P(x < 700) = 2.01
P(Z < 2.01) = 0.9778
P(x < 850) - P(x < 700) =
P(Z < 2.01) - P(Z < - 0.327)
0.9778 - 0.3718
= 0.606
Answer:5. 291503
Step-by-step explanation:
√28
2√7
5. 291503
Answer:
12g8h10
Multiply each variable quantity by 2 and combine.
<span> When you want the x </span>intercepts<span> (x,0):. Set y=0 then solve for x</span>
