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3 years ago

A researcher claiming that females were more empathetic than males would test that hypothesis by using __________ statistics.

2 answers:

6 0

<span>A researcher claiming that females were more empathetic than males would test that hypothesis by using inferential statistics.</span>

sergeinik [125]3 years ago

6 0

inferential is the correct answer because it goes with it and because my uncle is here and he taught me this but it really is the correct answer

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Column A is in the x-axis, and column B is on the y-axis. Which titles should replace A and B

pogonyaev

Column A: x-axis, input, domain

Column B: y-axis, output, range

Those are other ways to describe them

hope i helped:)

8 0

3 years ago

A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V

yan [13]

<h2>Answer:</h2>

(a) Attached to the response as Figure 1.

(b) 35.0Ω

Across 10.0Ω = 2.6Ω

Across 20.0Ω = 5.2Ω

<h2>Explanation:</h2>

(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.

(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e

R = R₁ + R₂ + R₃ -------------------(i)

Where;

R₁ = 5.0 Ω

R₂ = 10.0 Ω

R₃ = 20.0 Ω

<em>Substitute these values into equation (i) as follows;</em>

∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω

∴ R = 35.0 Ω

Therefore, the equivalent resistance is ∴ R = 35.0Ω

(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;

i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.

ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;

V = I R ------------------(ii)

Where;

V = Voltage supplied to the circuit = 9.0V

I = Current through the circuit

R = Resistance of the equivalent resistor = 35.0Ω

Substitute these values into equation (ii)

9.0 = I x 35.0

I = $\frac{9.0}{35.0}$

I = 0.26A

This is also the current flowing through each of the resistors separately.

iii. Calculate the voltage drop across

1.<em> 5.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 5.0Ω resistor

I = current through the 5.0Ω resistor = 0.26A

R = resistance of the 5.0Ω resistor = 5.0Ω

=> V = 0.26 x 5.0

=> V = 1.3V

2.<em> 10.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 10.0Ω resistor

I = current through the 10.0Ω resistor = 0.26A

R = resistance of the 10.0Ω resistor = 10.0Ω

=> V = 0.26 x 10.0

=> V = 2.6V

3.<em> 20.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 20.0Ω resistor

I = current through the 20.0Ω resistor = 0.26A

R = resistance of the 20.0Ω resistor = 10.0Ω

=> V = 0.26 x 20.0

=> V = 5.2V

7 0

3 years ago

The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T

ikadub [295]

Answer:

a) $a_{1}=3.7 m/s^{2}$

b) $a_{2}=3.68 m/s^{2}$

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

$v_{f}^{2}=v_{i}^{2}+2a\Delta x$

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.5}$

$a_{1}=3.7 m/s^{2}$

positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.6}$

$a_{2}=3.68 m/s^{2}$

positive acceleration.

I hope it helps you!

8 0

3 years ago

What is electrical energy

creativ13 [48]

Explanation:

it is energy that flows of electric charge it has the ability to do work or apply force to move an object.

8 0

3 years ago

Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,

REY [17]

Answer:

a) 238U, 40K and 87Rb, b) 235U and to a lesser extent 40K , c) he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0

3 years ago