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3 years ago

11

Okay whats 4+4 ok thats a question now this is my next question whoever opens this can you please report every question this dud

e posted because hes trolling his name is griffin something heres his profile https://brainly.com/app/profile/29647874/answers

2 answers:

MaRussiya [10]3 years ago

4 0

Answer:

ok, thanks for heads up

Step-by-step explanation:

grigory [225]3 years ago

4 0

Answer:

8

Step-by-step explanation:

Addition is one of the four basic operations of arithmetic, the other three being subtraction, multiplication and division. The addition of two whole numbers results in the total amount or sum of those values combined. The example in the adjacent image shows a combination of three apples and two apples, making a total of five apples. This observation is equivalent to the mathematical expression "3 + 2 = 5". - Wikipedia

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Do the data in the table represent a direct variation or inverse variation? Write an equation to model the data in the table?

notsponge [240]

Based on the given data; the table represent an inverse variation and the equation that model the data in the table is y = 8/x

<h3>Variation</h3>

Direct variation

y = k × x

4 = k × 2

4 = 2k

k = 4/2

k = 2

when y = 2 and x = 4

y = k × x

= 2 × 4

y = 8

Indirect variation

y = k/x

4 = k / 2

8 = k

when y = 2 and x = 4

y = k/x

2 = k/4

2 × 4 = k

k = 8

So,

y = k/x

y = 8/x

Learn more about variation:

brainly.com/question/6499629

#SPJ1

5 0

2 years ago

Find all complex solutions of 7x^2-7x+3=0

Sedaia [141]

Answer:

The complex solutions are

$x = (\frac{7-\sqrt{35} i }{42} , ) x = \frac{7 + \sqrt{35} i }{42})$

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given equation 7 x² - 7 x +3=0

$x = (\frac{-b -\sqrt{b^{2}-4ac } }{2ac} , \frac{-b +\sqrt{b^{2}-4ac } }{2ac})$

Given standard equation

a x² +b x +c =0

a = 7 , b= -7 and c=3

<u><em>Step(ii):-</em></u>

<u><em></em></u> $x = (\frac{-(-7) -\sqrt{(-7)^{2}-4(7)(3) } }{2(7)(3)} , \frac{-(-7) +\sqrt{(-7)^{2}-4(7)(3) } }{2(7)(3)})$ <u><em></em></u>

<u><em></em></u> $x = (\frac{7-\sqrt{(49-84 } }{42} , \frac{7 +\sqrt{(49-84 } }{42})$ <u><em></em></u>

$x = (\frac{7-\sqrt{35i^{2} } }{42} , \frac{7 + \sqrt{35i^{2} } }{42})$

$x = (\frac{7-\sqrt{35} i }{42} , \frac{7 + \sqrt{35} i }{42})$

<u><em></em></u>

8 0

2 years ago

Please help asap for a math test

Kruka [31]

Answer:

negative

positive

negative

positive

Step-by-step explanation:

-0.25

1/10

-0.2

1/2

5 0

2 years ago

The price of a computer was decreased by 30% to £147. What was the price before the decrease?

Sunny_sXe [5.5K]

Answer:

£210

Step-by-step explanation:

A decrease of 30% represents 70% of the original cost.

100% represents the original cost

Divide the price by 70 to find 1% then multiply by 100 for original cost

original cost = $\frac{147}{70}$ × 100 = 2.1 × 100 = £210

4 0

3 years ago

Read 2 more answers

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago