All categories

3 years ago

PLEASE HELP PICTURE PROVIDED

Mathematics

2 answers:

Nataly_w [17]3 years ago

8 0

The answer is A. $6.50

Tamiku [17]3 years ago

4 0

Answer:

the answer is A 6.50

Step-by-step explanation:

becasue u just have to look at the line and find the one hour

You might be interested in

What is 9/10-6/10 or can you not do it?

Olin [163]

Answer:

3/10

Step-by-step explanation:

9-6=3

both fractions have the same denominator so they keep it

9/10 - 6/10 = 3/10

5 0

2 years ago

In 2000, the total population of the U.S. was 281.4 million people. In 2010, it was 308.7 million people. What is the average ra

kifflom [539]

Answer:

2.73 million people per year

Step-by-step explanation:

Turn sentences into coordinates/points:

In 2000, 281.4 million people

(x₁, y₁) = (2000, 281.4)

In 2010, 308.7 million people

(x₂, y₂) = (2010, 308.7)

Rate of population per year:

$\sf \dfrac{y_2 - y_1}{x_2- x_1} \ \ where \ (x_1 , \ y_1), ( x_2 , \ y_2) \ are \ points$

$\rightarrow \sf \dfrac{308.7-281.4}{2010-2000}$

$\rightarrow \sf 2.73 \ million \ people$

5 0

1 year ago

3 different numbers which are all equivalent to 10

Zolol [24]

Answer:

2 4 6

Step-by-step explanation:

3 0

3 years ago

Use technology and the given confidence level and sample data to find the confidence interval for the population mean mu. Assume

Sphinxa [80]

Answer: (1.55, 6.45)

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline\ {x}\pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Significance level : $\alpha: 1-0.99=0.01$

Critical value : $z_{\alpha/2}=2.576$

Sample size : n=41

Sample mean : $\overline{x}=4.0\text{ kg}$

Standard deviation : $\sigma=6.1\text{ kg}$

Then, 99% confidence interval for population mean will be :_

$4\pm\ (2.576)\dfrac{6.1}{\sqrt{41}}\\\\\approx4\pm2.45\\\\=(4-2.45, 4+2.45)=(1.55, 6.45)$

8 0

3 years ago

Is KLMN below a rectangle?

8_murik_8 [283]

Should be C right? A rectangle demands that the diagonals are congruent

6 0

3 years ago