Answer: Variables

F and C are placeholders for numbers. They are called variables because they are allowed to vary, or change. If you change C then it affects F, and vice versa. If the values for the placeholders is not allowed to change, yet it holds a number, then it is considered a constant. In this case, we don't have constants or else the formula isn't too useful.

3 0

3 years ago

-0.000167 in scientific notation

tiny-mole [99]

$1.67 \times{10}^{ - 4}$

6 0

3 years ago

Sylvie finds the solution to the system of equations by graphing. y = x + 1 and y = x – 1 Which graph shows the solution to Sylv

Furkat [3]

Graphs of y = x+- 1 and of y = x - 1 are shown in the figure below.
The intersection of the two graphs is the solution for the two equations.

Answer:
There is no solution.

Explanation:
It is clear that the two lines are parallel and will never intersect.

Note that the equation for a straight line is of the form
y = mx + b
where
m = slope,
b = y-intercept.

y = x + 1 has slope of m=1.
y = x - 1 has slope of m=1.
The two slopes are equal.
Because parallel lines do not intercept, there is no solution.

6 0

3 years ago

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in,and a standard deviation given by

Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$ , then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$ . So, the probability density function is given by

$g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}$ , and

$P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx$ = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

4 0

3 years ago