Round 545,999 to the nearest hundred thousand

A landscape supply business charges $35 to deliver mulch. The cost of the mulch is $29.

Scorpion4ik [409]

Answer:

$209

Step-by-step explanation:

we know that

The equation of the line in slope intercept form is equal to

$y=mx+b$

where

m is the slope

b is the y-coordinate of the y-intercept (initial value)

Let

x ----> the volume of mulch in cubic yards

y ----> the total cost in dollars

we have that

The slope of the linear equation is

$m=29\ \frac{\$}{yd^3}$

$b=\$35$ ----> flat fee to deliver mulch

The linear equation is

$y=29x+35$

For x=6 yd^3

substitute in the equation and solve for y

$y=29(6)+35$

$y=\$209$

4 0

2 years ago

What is the value of x in the figure below? In this diagram, ABD~CAD

olga55 [171]

Answer:

x = 25/4

Step-by-step explanation:

Because of the known similarity of the triangles, we know that

10 x

----- = ----

16 10

Cross-multiplying, we get 16x = 100, and thus x = 100/16 = 50/8 = 25/4

x = 25/4

4 0

2 years ago

A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr

erastovalidia [21]

Answer:

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

$p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034$

The estimated standard error is:

$\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316$

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

$LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096$

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

8 0

2 years ago

4400 dollars is placed in an account with an annual interest rate of 8.25%.how much will be in the account after 22 years, to th

iVinArrow [24]

Answer:

Future Value = $2,516,900 cent

Step-by-step explanation:

Here, we're to calculate future values;

Given the following parameters

Investment = $4,400

Interest Rate = 8.25%

Time = 22 years