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2 years ago

!!!BRAINLIEST!!!

Mathematics

2 answers:

Anarel [89]2 years ago

7 0

I believe it is 60% I got you sis:))

kati45 [8]2 years ago

4 0

Answer: 60%

From what I learned, take this example from a pair of dice: If you want to know the probability of getting 3, you take the event (the 3 or the thing you want to know) and divided it by the outcome (or total)

in other words : probility = event(3)/outcome(6)

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Simplify the following expression.

adoni [48]

$$ 12 - 5[(4 + 2)\div\2 - (22 + 10)]= 157 $$
238 this answer

7 0

2 years ago

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Help ASAP i need help

Reptile [31]

Answer:

Isosceles Right Triangle

Step-by-step explanation:

We look at the sides first. Two sides are equal, which means that the triangle is isosceles (equilateral means three equal sides, and scalene means no sides equal). Then we look at the only angle indicated, which is the right angle. Thus, we come to the conclusion that the triangle is an Isosceles Right Triangle.

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2 years ago

Solve the equation<br><br> c/7=6 1/7

Softa [21]

c/7 = 6 1/7 ⇒ c = 43 ⇒ 43/7 = 6 1/7

Convert mixed fraction into a simple fraction

6 1/7 ⇒ (6 * 7 + 1)/7 = 43/7

c/7 = 43/7

proportionality:

a/b = c/d where ad = bc

a = c
b = 7
c = 43
d = 7

ad = bc ⇒ c7 = 7*43 ⇒ c7 = 301 ⇒ 7c/7 = 301/7 ⇒ c = 43

5 0

2 years ago

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4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$