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2 years ago

WILL GIVE BRAINLIEST IF YOU KNOW THE ANSWER!!! PLS HELP :)

Mathematics

1 answer:

galben [10]2 years ago

8 0

Answer:

A. 5.0 * 10⁻⁹ J

Step-by-step explanation:

Q is charge.

ΔV is electric potential energy.

All we have to do is plug both values in.

PE = (1/2)(1.0 * 10⁻¹⁰)(100) = (1/2)(1.0* 10⁻⁸) = 5.0 * 10⁻⁹ J

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Which of the following is equivalent to log34

olasank [31]

hope it helps u...

plz mark as brainliest...

4 0

3 years ago

I need help with the last two !!!

Dmitriy789 [7]

Step-by-step explanation:

G(x) = - x^2 + 10x

a + G(a) = a - a^2 + 10a = - a^2 + 11a or 11a - a^2 or a(11 - a)

1/G(a) = 1/(10a - a^2) = 1/a(10 - a)

6 0

2 years ago

In a random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand. C

horrorfan [7]

Answer:

90% confidence interval for the proportion of fans who bought food from the concession stand

(0.5603,0.6529)

Step-by-step explanation:

Step(i):-

Given sample size 'n' =300

Given data random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand.

Given sample proportion

$p^{-} = \frac{x}{n} = \frac{182}{300} =0.606$

level of significance = 90% or 0.10

Z₀.₁₀ = 1.645

90% confidence interval for the proportion is determined by

$(p^{-} - Z_{0.10}\sqrt{\frac{p(1-p)}{n} } , p^{-} +Z_{0.10}\sqrt{\frac{p(1-p)}{n} })$

$(0.6066 - 1.645\sqrt{\frac{0.6066(1-0.6066)}{300} } ,0.6066+1.645\sqrt{\frac{0.6066(1-0.6066)}{300} })$

(0.6066 - 0.0463 ,0.6066 + 0.0463)

(0.5603,0.6529)

final answer:-

90% confidence interval for the proportion of fans who bought food from the concession stand

(0.5603,0.6529)

5 0

3 years ago

Suppose that 45% of people have dogs. If two people are randomly chosen, what is the probability that they both have a dog

mylen [45]

Answer:

$P(Dogs) = 0.2025$

Step-by-step explanation:

Given

$Proportion, p = 45\%$

Required

Probability of two people having dog

First, we have to convert the given parameter to decimal

$p = \frac{45}{100}$

$p = 0.45$

Let P(Dogs) represent the required probability;

This is calculated as thus;

P(Dogs) = Probability of first person having a dog * Probability of second person having a dog

$P(Dogs) = p * p$

$P(Dogs) = 0.45 * 0.45$

$P(Dogs) = 0.45^2$

$P(Dogs) = 0.2025$

Hence, the probability of 2 people having a dog is  $P(Dogs) = 0.2025$ 

4 0

3 years ago

Consider the system of equations.

jok3333 [9.3K]

Given:

The system of equations:

$x-3y=9$

$\dfrac{1}{5}x-2y=-1$

To find:

The number that can be multiplied by the second equation to eliminate the x-variable when the equations are added together.

Solution:

We have,

$x-3y=9$ ...(i)

$\dfrac{1}{5}x-2y=-1$ ...(ii)

The coefficient of x in (i) and (ii) are 1 and $\dfrac{1}{5}$ respectively.

To eliminate the variable x by adding the equations, we need the coefficients of x as the additive inverse of each other, i.e, a and -a So, a+(-a)=0.

It means, we have to convert $\dfrac{1}{5}$ into -1. It is possible if we multiply the equation (ii) by -5.

On multiplying equation (ii) by -5, we get

$-x+10y=5$ ...(iii)

On adding (i) and (iii), we get

$7y=14$

Here, x is eliminated.

Therefore, the number -5 can be multiplied by the second equation to eliminate the x-variable.

6 0

3 years ago