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3 years ago

.16 divided by .03220 round to the nearest tenth

Mathematics

2 answers:

vredina [299]3 years ago

6 0

Answer:

Aosvsixixbs ss

Step-by-step explanation:

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nasty-shy [4]3 years ago

6 0

Answer:

0.2

Step-by-step explanation:

03220 divided by .16 is 0.20125. Rounded to the nearest tenth it is 0.2

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To divide by a decimal multiply both the dividend and divisor by the same number so that it eliminates the decimal in the divisor - it is usually easiest to multiply by a power of 10.

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A movie theater complex gave a poll as viewers exited from two theaters showing the same film. They had a 20% response rate from

sp2606 [1]

Answer:

Let´s define the variables:

x = number of people in theater A

y = number of people in theater B

We know that a 20% (or 0.2 in decimal form) of the people in theater A answered the survey, then the number of people from theater A that answered the survey is: Na = 0.2*x

We know that a 30% (or 0.3 in decimal form) of the people in theater B answered the survey, then the number of people from theater B that answered the survey is: Nb = 0.3*y

There where 800 people in total (counting from the two theatres)

Then x + y = 800.

A) In this case we have:

Na = Nb

this means that:

0.2*x = 0.3*y

and we also have the equation:

x + y = 800

To solve this, we need to isolate one variable in one of the equations, i will isolate x in the second equation:

x = 800 - y

And now we can replace this in the other equation to get:

0.2*(800 - y) = 0.3*y

We need to solve this for y:

160 - 0.2*y = 0.3*y

160 = 0.3*y + 0.2*y = 0.5*y

160/0.5 = y = 320

Then there where 320 people in theater B, and:

x + y = 800

x + 320 = 800

x = 800 - 320 = 480

There where 480 people in theather A.

B) Now we have:

x = 440

y = 360

then:

Na = 0.2*440 = 88

In theater A, 88 people answered the survey.

Nb = 0.3*360 = 108

In theater B, 108 people answered the survey.

In total, 88 + 108 = 196 people answered the survey.

5 0

2 years ago

According to Lambert's law, the intensity of light from a single source on a flat surface at point P is given by Upper L equal

malfutka [58]

Answer:

(a) $L = k*(1 - sin^{2}(\theta))$

(b) L reaches its maximum value when θ = 0 because cos²(0) = 1

Step-by-step explanation:

Lambert's Law is given by:

$L = k*cos^{2}(\theta)$ (1)

(a) We can rewrite the above equation in terms of sine function using the following trigonometric identity:

$cos^{2}(\theta) + sin^{2}(\theta) = 1$

$cos^{2}(\theta) = 1 - sin^{2}(\theta)$ (2)

By entering equation (2) into equation (1) we have the equation in terms of the sine function:

$L = k*(1 - sin^{2}(\theta))$

(b) When θ = 0, we have:

$L = k*cos^{2}(\theta) = k*cos^{2}(0) = k$

We know that cos(θ) is a trigonometric function, between 1 and -1 and reaches its maximun values at nπ, when n = 0,1,2,3...

Hence, L reaches its maximum value when θ = 0 because cos²(0) = 1.

I hope it helps you!

5 0

3 years ago

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There are three choices of jellybeans: grape, cherry and orange. If the probability of getting a grape is 3/10 and the probabili

melisa1 [442]

can u plzz explain me the question

7 0

3 years ago

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a 4.50 g coin of copper absorbed 54 calories of heat. what was the final temperature of the copper if the initial temperature wa

I am Lyosha [343]

The final temperature of copper is 155.43°C.

Step-by-step explanation:

The formula to apply is Q=m*Δt*c where ;

Q=heat energy, m=mass, c=specific heat capacity and Δt= change in temperature

Given that Q=54 calories , m=4.50 g and c=0.092 cal/g°C

In this question find Δt then ,the final temperature.

Δt=Q/m*c

Δt=54/4.50*0.092 =130.43°C

To find final temperature, apply the formula;

Δt=Final temperature - Initial temperature

130.43°C=Final temperature-25°C

130.43°C+25°C=Final temperature

155.43°C

Learn More

Specific Heat Capacity : brainly.com/question/12431744

Keyword : specific heat capacity

#LearnwithBrainly

3 0

3 years ago

70 points !!! So try

Kazeer [188]

Answer:

1/ 5²

Step-by-step explanation:

= 5³ × 1/5⁵

= 125 × 1/5⁵

= 125/3125

= 1/25 or 1 / 5²

3 0

3 years ago

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