All categories

4 years ago

The slope of a line is?

Mathematics

1 answer:

valkas [14]4 years ago

6 0

Answer:

The correct answer is d) the ratio of the change in y to the change in x along the line.

We can tell this by looking at the slope formula.

m (slope) = (y2 - y1)/(x2 - x1)

You might be interested in

48/31 in simplest form

Eduardwww [97]

48/31 is already at its simplest form, however you can change it to a mixed fraction which would become 1 17/31

3 0

4 years ago

Read 2 more answers

As a salesperson, you are paid $50 per week plus $3 per sale. This week you want your pay to be at least $110. What is the minim

alexgriva [62]

110 (goal) - 50 (base pay) = 60 (pay from sales only)

60 (total pay from sales) / 3 (cost per individual sale) = 20 (qty of sales)

7 0

3 years ago

Read 2 more answers

A jar contains 8 large red marbles, 5 small red marbles, 9 large blue marbles, and 7 small blue marbles. If a marble is chosen a

Eddi Din [679]

Answer:

P(blue|small) > P(small|blue)

Step-by-step explanation:

P(blue|small): 7 / (5 + 7) = 7 / 12

P(small|blue): 7 / ( 9 + 7) = 7 / 16

7 / 12 > 7 / 16

4 0

3 years ago

What is the quotient of 2.408 • 10^24 divided by 6.02 • 10^23

Illusion [34]

Answer:

2.408

Step-by-step explanation:

8 0

3 years ago

Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar

klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

$\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}$

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

$CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack$

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

$CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack$

Where (from tables):

$Z_{0.99}=2.33$

Finally, the interval at 98% confidence level is:

$CI(\mu)=\lbrack28.94,31.06\rbrack$

4 0

1 year ago