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2 years ago

Plz solve . 3+ x > -9

Mathematics

1 answer:

STALIN [3.7K]2 years ago

5 0

Answer:

Step-by-step explanation:

3 + x>-9 x=12 15>-9

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Admission for adults is 12 dollars. Admission for kids is eight dollars.One day 575 people entered the cave paying 5600 how many

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Answer:

Step-by-step explanation:

Admission for adults is 12 dollars. Admission for kids is eight dollars.One day 575 people entered the cave paying 5600 how many adults into the cave

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Answer:

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What is the following quotient 9+sqrt 2/4-sqrt 7

babunello [35]

9 + √2 / 4 - √7

Multiply the numerator and denominator by the conjugate.

Exact form:

(36 + 9√ 7 + 4√2 + √14) / 9

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3 years ago

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How to find equation hyperbola with asymptotes, center and transverse?

lapo4ka [179]

The equation for a hyperbola is (x-h)/a - (y-k)/b = 1
Or (y-k)/a - (x-h)/b = 1
h represents the x value of the coordinate
k value represents the y value of the coordinate
together they represent a point, which is the center
So (h,k) is (x,y)

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3 years ago

A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow

Mrrafil [7]

Answer:

$\displaystyle \frac{54}{5405}$ .

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

$\displaystyle \left(50\atop 5\right) = 2,118,760$ .

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

$\displaystyle \left( 28\atop 2\right) = 378$ .

Number of ways to choose 3 green candies out of a batch of 8:

$\displaystyle \left(8\atop 3\right)=56$ .

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

$\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168$ .

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

$\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}$ .

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3 years ago