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3 years ago

14

You pay $2.40 for 3/4 pounds of grapes . Your friend buys a pound of grapes at the same time and estimates that he will pay $1.8

0. What can you say about your friends estimate ?

1 answer:

Lera25 [3.4K]3 years ago

7 0

You can say that they would have enough for about half a pound alittle more

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Please help i will mark you brainliest !

pav-90 [236]

Answer:

Take two points
y=-2x-3

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3 years ago

Is this answer are right, please help

uranmaximum [27]

Answer:

Your answers are correct. However, the instructions say to write the formula, and in my class you would write A= bh ÷ 2.

Step-by-step explanation:

However, the instructions say to write the formula, and in my class you would write A= bh ÷ 2. Also, you may want to write A= for every line of math that you do. If your class doesn't do that, then disregard that. :)

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3 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

4 years ago

Read 2 more answers

What is the solution set of {x | x < -3} ∩ {x | x > 5}

lozanna [386]

Any $x$ in this set will be real numbers that are both less than $-3$ and greater than $5$ . But that's not possible, so this set is empty.

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3 years ago

Find the volume of this rectangular pyramid.

Ivahew [28]

Volume is just length x width x height. 5 x 6 x 7 is 210. 210cm

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2 years ago