A group of 5 friends decides to share the cost of a party together. If the party costs $1,100, how much does each friend pay?

Why do you think a removable discontinuity (hole) doesn't produce an asymptote on the graph of a polynomial function, even thoug

Monica [59]

Because removable discontituity means that the limit of the function at that point has a finite value, and then you define the value of the function as that valu (the limit value).

An asymptote means that the limit of the function goes to positive or negative infinity.

You cannot meet both conditions, finite and infinity limit at the same time.

8 0

1 year ago

The average American man consumes 9.8 grams of sodium each day. Suppose that the sodium consumption of American men is normally

Alex Ar [27]

Answer:

(a) The distribution of X is N (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 X is defined as the amount of sodium consumed.

The random variable X has an average value of, μ = 9.8 grams.

The standard deviation of X is, σ = 0.8 grams.

(a)

It is provided that the sodium consumption of American men is normally distributed.

The random variable X follows a normal distribution with parameters μ = 9.8 grams and σ = 0.8 grams.

Thus, the distribution of X is N (9.8, 0.8²).

(b)

If X ~ N (µ, σ²), then $Z=\frac{X-\mu}{\sigma}$ , 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 (X) to z-scores.

Compute the probability that an American consumes between 8.8 and 9.9 grams of sodium per day as follows:

$P(8.8$

$=P(-1.25$

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:

$P(x_{1}$

Compute the value of z as follows:

$P(-z$

The value of z for P (Z < z) = 0.65 is 0.39.

Compute the value of x₁ and x₂ as follows:

$-z=\frac{x_{1}-\mu}{\sigma}\\-0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8-(0.39\times 0.8)\\x_{1}=9.488\\x_{1}\approx9.5$ $z=\frac{x_{2}-\mu}{\sigma}\\0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8+(0.39\times 0.8)\\x_{1}=10.112\\x_{1}\approx10.1$