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SCORPION-xisa [38]

3 years ago

12

You have $6 to spend on lip balm and hand sanitizer. The equation $1.5x+2.5y=6$ represents this situation, where $x$ is tubes of

lip balm and $y$ is bottles of hand sanitizer. How many tubes of lip balm can you buy when you do not buy any bottles of hand sanitizer?

1 answer:

galben [10]3 years ago

5 0

Answer:4

Step-by-step explanation:

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The owner of a bike shop sells unicycles and bicycles and keeps inventory by counting seats and wheels. One day, she counts 15 s

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Im going to guess its either b or c.

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Find the distance between these points. <br> A(5,8), B(-3,4)

konstantin123 [22]

Answer:

<u>Distance</u><u> </u><u>between</u><u> </u><u>the</u><u> </u><u>points</u><u> </u><u>is</u><u> </u><u>8</u><u>.</u><u>9</u><u>4</u><u> </u><u>units</u>

Step-by-step explanation:

General formula:

${ \boxed{ \bf{distance = \sqrt{ {(x_{1} - x _{2} )}^{2} + {(y _{1} - y _{2} ) }^{2} } }}}$

substitute:

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3 years ago

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Unite rate. 3 cans of soup for $0.99

soldi70 [24.7K]

Answer:

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Step-by-step explanation:

Unit rate=1

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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:

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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

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lukranit [14]

Answer:

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Step-by-step explanation:

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