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2 years ago

Are the compositions of f(x) = 1 and g(x) = 2 commutative? Why or why not?

Mathematics

1 answer:

AlladinOne [14]2 years ago

4 0

Answer:

first option

Step-by-step explanation:

They are cumulative because they are constant functions.

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Select the representation that does not change the location of the given point. (4, 110°)

miss Akunina [59]

Answer:

(4,470°)

Step-by-step explanation:

The representation that does not change the location of (4, 110°) in polar coordinates are all points that are coterminal with the given point.

The only point among the given options that is coterminal with (4, 110°) is (4, 470)°

The two points have the same magnitude and 470°-360°=110°.

Since 110° is coterminal with 470° and the two points have the same magnitude with the same sign, the two points represent the same location in polar coordinates.

The correct choice is (4,470°)

3 0

4 years ago

What is the value of x? 0.9(x+1.4)−2.3+0.1x=1.6

mash [69]

Expand

Simplify 0.9x + 1.26 - 2.3 + 0.1 * x to x - 1.04

add 1.04 to both sides

simplify 1.6 + 1.04 to 2.64

Answer: x = 2.64

5 0

3 years ago

Read 2 more answers

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of <em>z</em> for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

3 years ago

The sandbox at a playground is 2 feet tall.The area of the base of the sandbox mis 80 square feet.How much sand does it take to

DaniilM [7]

360 square feet of sand, I think

8 0

3 years ago

Jamal is planting seeds for a garden nursery.He plants 9 seeds in each container.If Jamal has 296 seeds to plant , about how man

ludmilkaskok [199]

He will use 33 containers because 9 goes into 296 32.8 times but you cant have .8 of a container so you round it up to 33

8 0

3 years ago