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3 years ago

Miguels employer pays $1,825 in health insurance and 593 in life insurance per year. He also gets $2,860 in paid time off per

Mathematics

1 answer:

Luden [163]3 years ago

7 0

I think that it’s c 58.668 I don’t know I’m really don’t sure I wish you the best of luck

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What happens to the force between two charged objects when you triple the magnitude of both charges?

Vika [28.1K]

If the force between two charged objects when you triple the magnitude of both charges. The Force will become 9 times the initial force.

<h3>What is coulombs law?</h3>

According to coulombs law force between two charges is given by

$F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}$

Here, R is the distance between both the charges Q and q.

Let the force on the charges be F1

The distance of separation = r

The magnitudes of the charges q1 and q2

K = Coulumb's constant

$F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}$

Let the force on the charges be F

The distance of separation = r

The magnitudes of the charges 3q1 and 3q2

K = Coulumb's constant

So,

$F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}$

$F = \dfrac{1}{4\pi \epsilon }\dfrac{3Q\times 3q}{r^2}\\\\F = \dfrac{1}{4\pi \epsilon }\dfrac{9Qq}{r^2}\\\\F =9 \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}\\\\F = 9F_1$

Learn more about coulombs law;

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8 0

2 years ago

Which ordered pairs are in a proportional relationship with (0.2, 0.3)?

MAXImum [283]

Answer:

C and E

Step-by-step explanation:

We are given that

$x_1=0.2,y_1=0.3$

$\frac{y}{x}=\frac{0.3}{0.2}=\frac{3}{2}$

$k=\frac{y}{x}=\frac{3}{2}$

A. $x_2=1.2,y_2=2.3$

$\frac{y_2}{x_2}=\frac{2.3}{1.2}=\frac{23}{12}\neq=\frac{3}{2}$

Hence, it is not in proportional relationship with (0.2,0.3)

B.(2.7,4.3)

$x_3=2.7,y_3=4.3$

$\frac{y_3}{x_3}=\frac{4.3}{2.7}=\frac{43}{27}\neq\frac{3}{2}$

Hence, it is not in proportional relationship with (0.2,0.3).

C.(3.2,4.8)

$x_4=3.2,y_4=4.8$

$\frac{y_4}{x_4}=\frac{4.8}{3.2}=\frac{3}{2}$

Hence, the ordered pair (3.2,4.8) are in a proportional relationship with (0.2,0.3).

D.(3.5,5.3)

$\frac{y_5}{x_5}=\frac{5.3}{3.5}=\frac{53}{35}\neq \frac{3}{2}$

Hence, the ordered pair (3.5,5.3) are not in a proportional relationship with (0.2,0.3).

E.(5.2,7.8)

$\frac{y_6}{x_6}=\frac{7.8}{5.2}=\frac{3}{2}$

Hence, the ordered pair (5.2,7.8) are in a proportional relationship with (0.2,0.3).

4 0

2 years ago

Ask Your Teacher Consider a binomial random variable with n = 5 and p = 0.8. Let x be the number of successes in the sample. Eva

Readme [11.4K]

Answer:

0.942 is the required probability.

Step-by-step explanation:

We are given the following in the question:

x is a binomial random variable with n = 5 and p = 0.8.

Then,

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

We have to evaluate:

$P(x \geq 3) = P(x = 3) + P(x = 4) +P(x=5)\\\\= \binom{5}{3}(0.8)^3(1-0.8)^2 +\binom{5}{4}(0.8)^4(1-0.8)^1 +\binom{5}{5}(0.8)^5(1-0.8)^0\\\\= 0.2048 +0.4096+0.3276=0.942$