Which algebraic expression below represents four multiplied by a number?

Delia is a graphic designer who charges an hourly rate. She made the graph to show her total earnings for various numbers of hou

stich3 [128]

Answer:

approximately 7 hours

see the explanation

Step-by-step explanation:

The picture of the question in the attached figure

Let

x ----> the time worked in hours

y ----> the amount earning in dollars

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line <u><em>and the line passes through the origin</em></u>

so

In this problem we have a proportional relationship

step 1

Find approximately how many hours would it take Delia to earn $250 ?

Looking at the graph

For y=$250 The value of x is approximately 7 hours

see the attached figure N 2

step 2

Find the exact value

from the graph take the point (5,175)

Find the value of k

$k=\frac{y}{x}$ ----> $k=\frac{175}{5}=\$35\ per\ hour$

The linear equation is

$y=35x$

For y=$250

substitute

$250=35x$

solve for x

$x=7.14\ hours$

8 0

3 years ago

How many triangles exist with the given side lengths of 3cm, 5cm, and 9cm?

Irina-Kira [14]

Only one triangle exist

3 0

4 years ago

Read 2 more answers

A study reveals that 65% of homeowners in Omaha own a lawn mower. To test this claim, you form a sample of 500 homeowners in Oma

kotegsom [21]

Answer:

There is no enough evidence to reject the claim using a level of significance of 5%

Step-by-step explanation:

Let p be the true proportion of homeowners in Omaha who own a lawnmower. We want to test

$H_{0}: p = p_{0} = 0.65$ vs $H_{1}: p\neq 0.65$ (two-tailed alternative).

We have a large sample of size n = 500 and a point estimate for p is $\hat{p}=0.68$ . The test statistic is given by

$Z=\frac{\hat{p}-p_{0}}{\sqrt{p_{0}(1-p_{0})/n}}=\frac{0.68-0.65}{\sqrt{0.65(1-0.65)/500}}$ where $\sqrt{0.65(1-0.65)/500}$ is the standard error of $\hat{p}$ when $H_{0}$ is true. The observed value is $z_{0}$ = 1.4064 which comes from a standard normal distribution. Because we want to use a level of significance of 5%, our rejection region is RR={z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles respectively. Because 1.4064 does not belong to RR, we fail to reject the null hypothesis.