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3 years ago

I NEED HELP!! Pls help

Mathematics

1 answer:

Mila [183]3 years ago

4 0

Answer:

Number 4

Step-by-step explanation:

A theoretical probability chance is a probably where all the other chances are equal. In this case, you can roll 6 different types of numbers on the dice: 1, 2, 3, 4, 5, and 6. And the number of times the dice was rolled was 60 times. So, we can divide 60/6 because there were 60 rolls and 6 different possibilities. 60/6=10, so 10 is the theoretical probability. As you can see, the graph shows that the number 4 was rolled 10 times, which means that number 4 is the experimental probability that matched the theoretical probability.

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A phone cost 199$ and is on sale for 25% off. What is the sale price?

andrey2020 [161]

Answer:

See below.

Step-by-step explanation:

Divide the percentage by 100 to turn it into decimal form.

$\frac{25}{100}=.25$

Then, multiply the percentage by the original amount.

$199*.25=49.75$

Last, subtract that from the original price.

$199-49.75=149.25$

Answer: $149.25$

:)

6 0

3 years ago

Read 2 more answers

Two cars leave towns 760 kilometers apart at the same time and travel toward each other. One car's rate is 12 kilometers per hou

zaharov [31]

Answer:

89 km/hr

Step-by-step explanation:

Distance = Rate(speed) × Time

D = R × T

T = 4 hours

One car's rate is 12 kilometers per hour less than the other's.

Hence:

First's car's rate = r × 4 = 4r

Second car's distance = (r + 12) × 4 = 4r + 48

4r + 4r + 48 = 760

8r + 48 = 760

8r = 760 - 48

8r = 712

r = 712/8

r = 89 km/hr

The rate of the slower car is 89 km/hr

3 0

3 years ago

1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$