Write as a fraction and simplify 4^-4

Jasmine can run 4 5/6 miles in 2/3 of an hour. how many miles can she run 1 hour simplify comepletly

Vikentia [17]

Remark

the question depends on d= r * t

d = the distance in miles.

r = the rate of travel in miles per hour

t = the time in hours.

Step One

Rearrange the formula so you are solving for the rate.

d = r * t Divide by t

4 5/6 = (4*6 + 5)/6 = 29/6

r = d / t

Step Two

Substitute into the equation.

$r = \dfrac{\dfrac{29}{6}}{\dfrac{2}{3} }$

Step Three

Turn the bottom fraction upside down. Multiply these two together.

r = $\dfrac{29}{6} * \dfrac{3}{2}$

r = 29/4 miles hour.

Step Four

Solve for d when t = 1 hour.

d = ??

r = 29/4 miles / hour

t = 1 hour.

d = 29/4 * 1

d = 29/4 miles

That's as simple as you can get it unless you want 7 1/4 miles or 7.25 miles

7 0

3 years ago

Which graph represents the function y= 2x-4?sty すす0-3xits : :os : ;-32pr ftーVetーTT it tie。pttーT-it-T +--+-

Sergio039 [100]

Solutions

You have been given the y-intercept of -4. The meaning of this is that the given line will go through the point (0, -4). Plug in 1 for x and find y.

$
y = 2x - 4

y = 2 * 1 - 4

y = 2 - 4

y = -2$
<span>

</span>In graph C,= x = 1, y = -2

<span>Graph C is the correct answer.</span>

4 0

4 years ago

Read 2 more answers

Among all monthly bills from a certain credit card company, the mean amount billed was $465 and the standard deviation was $300.

Fynjy0 [20]

Answer:

0.02% probability that the average amount billed on the sample bills is greater than $500.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10$ .