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2 years ago

To summarize and analyze data with both a crosstabulation and charting, Excel typically pair

Physics

1 answer:

r-ruslan [8.4K]2 years ago

7 0

Answer:

To summarize and analyze data with both a crosstabulation and charting, Excel typically pair PivotCharts with PivotTables (option b)

Explanation:

Pivot Tables allows us to deal with a high volume of data. This kind of tables are mainly designed to summarize, analyze and explore data in a simple way. Meanwhile, Pivot Charts gives the possibility to visualize the summarizing data provided in PivotTables, enabling the detection of trends, correlations,comparisons and so on. Because of that Pivot tables and pivot charts are usefull and complementary tools of Excel.

Summarizing, the proper option is b.

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Jodi made a list about electric current to help her study for a test. 1) Movement of electrons is continuous in a current. 2) El

Vladimir79 [104]

Answer:

Rate at which current flows is measured in amperes

Explanation:

The rate of flow of electrons constitutes the current. The electrons flow from lower electric potential to higher electric potential. When there is no potential difference then no electron will flow. The direction of the current and the electron are in opposite direction.

The direction of electron from the negative terminal to the positive terminal. The direction of current is from the positive terminal to the negative terminal.The current is measured in ampere.

The expression for current and the charge is as;

$I=\frac{q}{t}$

Here, q is the charge, t is the time taken and I is the current.

According to the given problem, Jodi made a list about electric current to help her study for a test. He described that electrons move from areas of low to high electric potential, voltage causes current to flow and movement of electrons is continuous in a current.

But he did error. It should be "rate at which charges flow" instead of rate at which current flow.

Therefore, the option (4) is correct.

4 0

3 years ago

Read 2 more answers

In Speed Study Number 1, we looked at two cars traveling the same distance at different speeds on city streets. Car "A" traveled

Ganezh [65]

Answer:

230.4 s

Explanation:

The speed of car A is

$v_A = 35 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car A is

$t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h$

The speed of car B is

$v_B = 45 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car B is

$t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h$

So the difference in time is

$\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h$

Which corresponds to

$\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s$

so car B arrived 230.4 s before car A.

5 0

2 years ago

Some substances have the same chemical composition, but their atoms are arranged differently. A classic example is carbon. A pen

Bumek [7]

No, the density of diamond and graphite would not be the same

Explanation:

What is density?

Density is the amount of substance per unit volume. It is simply mass divided by volume.

Density is greatly influenced by the way substances are packed.

Substances that are well packed will have lower volume for the same amount of matter than another that is poorly packed.

The carbon atoms in graphite are poorly packed. They are arranged layers upon layers.
Diamond carbon atoms have a cross-linked networked pattern. They are well packed.
For the same mass of matter, graphite will take up more space than diamond.

Since:

Density = $\frac{mass}{volume}$

The one that has a lesser volume will have a higher density.

Therefore diamond will have a higher density.

learn more:

Density brainly.com/question/5055270

#learnwithBrainly

4 0

2 years ago

Read 2 more answers

We all depend on electricity. Most electricity is created by electromagnetic generators at large power plants and distributed th

Anuta_ua [19.1K]

Answer:

1) not so long (maybe an hour or two)

2) access to information through the internet will be most affected if my computer and mobile phone run out of battery power.

3) yes, one should prepare for power outage. This can be done by having a standby alternative source of power like the use of inverters that stores electrical energy in form of chemical energy, and small internal combustion engine powered electric generators.

4) solar panels can be used to draw power from incident sun rays, this power can be stored in an inverter for future use in case of a power outage.

5) energy from the sun is converted into direct current which is then supplied to an accumulator in the opposite direction to its flow of current. When the energy is needed, it can be used directly, or converted to an alternating current. This is achieved by connecting its terminal to the supply. Electric field is generated by flow of ions and electrons within the working chemical (e.g lithium).

Explanation:

8 0

2 years ago

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn

KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0

3 years ago