All categories

4 years ago

An unknown or changeable quantity is

Mathematics

1 answer:

PSYCHO15rus [73]4 years ago

5 0

Answer:

Dependent Variable

Step-by-step explanation:

For example, the number of different colors on the page cannot be changed, but driving 80 miles per hour is changeable.

You might be interested in

You want to know the opinions of dog owners across America on the issue of establishing a law requiring individuals who own pit

sergij07 [2.7K]

Answer:

The results of this survey may be affected by nonrepsonse bias

Step-by-step explanation:

Just took this test

8 0

3 years ago

Mrs. Bennett has graded 20% of her 150 students' papers. How many papers does she still need to finish grading?

saveliy_v [14]

120

You can find out how many she has graded by multiplying 150 by the decimal version of 20% (.20). Then, subtract the number you get by 150.

3 0

3 years ago

Read 2 more answers

what is the temperature in degrees celsius that is equivalent to a temperature of 40 degrees Fahrenheit

sdas [7]

4.4 in Celsius I’m pretty sure

3 0

3 years ago

Solving systems by elimination 3x + 4y equals -3 5x + 3y equals 6

taurus [48]

Answer:

Move all terms that don't contain x to the right side and solve.

x=−1 − 4y/3

Move all terms that don't contain x to the right side and solve.

x=6/5 − 3y/5

3 0

3 years ago

Read 2 more answers

A. Use the one-mean t-interval procedure with the sample mean, sample size, sample standard deviation, and confidence level

PtichkaEL [24]

Answer:

a) (17.227, 22.773)

b) 2.773

c) 2.773

Step-by-step explanation:

Given:

Sample size, n = 25

Standard deviation, s = 4

Sample mean, x' = 20

Level of significance, a = 0.98 = 1 - 0.98 = 0.02

The degrees of freedom, df, for a t-distribution = n - 1 = 25 - 1 = 24

Using the t table, the Critical value = $t_\alpha _/_2, _d_f = t_0_._0_2_/_2, _2_4 = t_0_._0_1_, _2_4 = 3.4668$

Margin of error, E = $t_\alpha _/_2, _d_f * \frac{\sigma}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773$

Limits of 98% confidence interval, we have:

Lower limit : x' - M.E = 20 - 2.773 = 17.227

Upper limit: x' + M.E = 20 + 2.773 = 22.773

Therefore, (17.227, 22.773) is 98% confidence interval.

b) Let's the margin of error by taking half the length of the confidence interval.

Since we are to use half the length of CI, we have:

$M.E = \frac{22.773 - 17.227}{2} = 2.773$

c) $M.E = t_\alpha _/_2, _d_f *s/\sqrt{n}$

$= t * \frac{s}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773$

4 0

3 years ago