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3 years ago

Evaluate

0%5Cfrac%7B6%7D%7B7%7D%20" id="TexFormula1" title="\frac{x}{y} for x = \frac{2}{3} and y = \frac{6}{7} " alt="\frac{x}{y} for x = \frac{2}{3} and y = \frac{6}{7} " align="absmiddle" class="latex-formula">
A $\frac{7}{9}$
B $\frac{8}{21}$
C $\frac{8}{10}$
D $\frac{12}{21}$

Mathematics

1 answer:

Whitepunk [10]3 years ago

4 0

$\displaystyle\\
\frac{x}{y} ~~~ \Big( \texttt{ for } x = \frac{2}{3} \texttt{ and } y = \frac{6}{7} \Big)= ?\\ \\ \\ 
\Longrightarrow~~~\frac{x}{y}= \frac{~\frac{2}{3} ~}{\frac{6}{7} }= \frac{2}{3} \cdot \frac{7}{6} =\frac{2\cdot 7}{3\cdot6 }= \frac{7}{3\cdot 3 }= \boxed{\frac{7}{9}} \\ \\ \\ 
\texttt{Correct ansver: A}$

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Step-by-step explanation:

All you have to do is combine the like terms.

Like terms are the terms that have the same variable and same exponent.

The like terms in this equation are 6x^2 and 2x^2, 11x and -17x, and -3 and -4.

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2 years ago

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

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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:

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

Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

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