Which of the following equations has the same solutions as the equation 3x+2=-12? Check all that apply.
1 answer:
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Answer:
(a) The distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).
(b) The probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.
(c) The middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.
Step-by-step explanation:
The random variable <em>X</em> is defined as the amount of sodium consumed.
The random variable <em>X</em> has an average value of, <em>μ</em> = 9.8 grams.
The standard deviation of <em>X</em> is, <em>σ</em> = 0.8 grams.
(a)
It is provided that the sodium consumption of American men is normally distributed.
The random variable <em>X</em> follows a normal distribution with parameters <em>μ</em> = 9.8 grams and <em>σ</em> = 0.8 grams.
Thus, the distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).
(b)
If X ~ N (µ, σ²), then , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z ~ N (0, 1).
To compute the probability of Normal distribution it is better to first convert the raw score (<em>X</em>) to <em>z</em>-scores.
Compute the probability that an American consumes between 8.8 and 9.9 grams of sodium per day as follows:
Thus, the probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.
(c)
The probability representing the middle 30% of American men consuming sodium between two weights is:
Compute the value of <em>z</em> as follows:
The value of <em>z</em> for P (Z < z) = 0.65 is 0.39.
Compute the value of <em>x</em>₁ and <em>x</em>₂ as follows:
Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.
Answer:
Width = 18, height = 3
Or
Width = 6, height = 9
Step-by-step explanation:
Complete question
( A piece of sheet metal 24 inches wide is bent to form a gutter as shown in the picture below, if the cross sectional area is 54 square inches, find the depth of the gutter).
X = width, Y = height
X - 2y = 24........ equation i
X = 24 - 2y.......... equation ii
Area = xy
54 = xy........ equation iii
Put equation ii into equation iii
54 = (24 - 2y) * y
54 = 24y - 2y²
2y² - 24y + 54 = 0........ equation iv
Solving equation iv using quadratic methods,
Factors = (y-3), (y-9)
(Y-3)(y-9) = 0
y = 3 or y = 9
Substitute y = 3 or y = 9 into equation ii
X = 24 -2(3) = 18 or x = 24 - 2(9) = 6
If Width = 18, height = 3
Or
Width = 6, height = 9
