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Free_Kalibri [48]
3 years ago
13

Four less than the product of one and a number 2.

Mathematics
1 answer:
zysi [14]3 years ago
6 0

Answer:

(1*2)-4  = -2

Step-by-step explanation:

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you do 77 divided by 14 you get 5.5 so the greatest number of teams is 5.

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Which of the following functions is graphed below?
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Quadratic

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True or false "No rectangle is a rhombus?"
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False.
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Find the length of the third side. If necessary, write in simplest radical form.
Arlecino [84]
Answer:

9

Explanation:

9 root 2 = root 81 x root 2 = root 162
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2 years ago
An urn contains two black balls and three red balls. If two different balls are successively removed, what is the probability th
lidiya [134]

You can extract two balls of the same colour in two different way: either you pick two black balls or two red balls. Let's write the probabilities of each pick in each case.

Case 1: two black balls

The probability of picking the first black ball is 2/5, because there are two black balls, and 5 balls in total in the urn.

The probability of picking the second black ball is 1/4, because there is one black ball remaining in the urn, and 4 balls in total (we just picked the other black one!)

So, the probability of picking two black balls is

P(\text{two blacks}) = \dfrac{2}{5} \cdot \dfrac{1}{4} = \dfrac{2}{20} = \dfrac{1}{10}

Case 2: two red balls

The probability of picking the first black ball is 3/5, because there are three red balls, and 5 balls in total in the urn.

The probability of picking the second red ball is 2/4=1/2, because there are two red balls remaining in the urn, and 4 balls in total (we just picked the other red one!)

So, the probability of picking two red balls is

P(\text{two reds}) = \dfrac{3}{5} \cdot \dfrac{1}{2} = \dfrac{3}{10}

Finally, the probability of picking two balls of the same colour is

P(\text{same colour}) = P(\text{two blacks})+ P(\text{two reds}) = \dfrac{1}{10} + \dfrac{3}{10} = \dfrac{4}{10} = \dfrac{2}{5}

7 0
3 years ago
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