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3 years ago

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The Green Goober, a wildly unpopular superhero, mixes 33 3 3 liters of yellow paint with 55 5 5 liters of blue paint to make 88

8 8 liters of special green paint for his costume.

1 answer:

Lera25 [3.4K]3 years ago

5 0

Check picture ................ z z z z

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10$ is spent on 2 big pails and 1 small one. A smail pail cosf half as much as each big one. How much is spent on a big pail? Ho

Mila [183]

Answer:

<u>$4</u> is spent on a big pail and <u>$2</u> is spent on a small pail.

Step-by-step explanation:

Given:

10$ is spent on 2 big pails and 1 small one.

A small pail cost half as much as each big one.

Now, to find the amount spent on a big pail and on a small pail.

Let the cost of a big pail be $x.$

And the cost of small pail be $\frac{x}{2} .$

As given, there are 2 big pails.

And the total money spent = $10.

According to question:

$(x+x)+\frac{x}{2}=10$

$2x +\frac{x}{2}=10$

$\frac{4x+x}{2} =10$

$\frac{5x}{2} =10$

<em>Multiplying both sides by 2 we get:</em>

$5x=20$

<em>Dividing both sides by 5 we get:</em>

$x=4.$

<em>The cost of a big pail</em> = $4.

<em>So, the cost of a small pail</em> = $\frac{x}{2} =\frac{4}{2}=\$2.$

Therefore, $4 is spent on a big pail and $2 is spent on a small pail.

6 0

3 years ago

Carmen wins $35 was in the contest. Tubes of her favorite brand of paint cost $4.80 each. Paint brushes cost $6.70 each. How man

Zanzabum

6.70 for paint brushes. for 2 is 6.70×2=$13.40

35-13.40=$21.60 left after 2 paint brushes

4.80×4=$19.20 for 4 tubes of paint

21.60-19.20=$2.40 left after 4 tubes of paint and 2 paint brushes

6 0

3 years ago

Suppose I ask you to pick any four cards at random from a deck of 52, without replacement, and bet you one dollar that at least

Tatiana [17]

Answer:

a) No, because you have only 33.8% of chances of winning the bet.

b) No, because you have only 44.7% of chances of winning the bet.

Step-by-step explanation:

a) Of the total amount of cards (n=52 cards) there are 12 face cards (3 face cards: Jack, Queen, or King for everyone of the 4 suits: clubs, diamonds, hearts and spades).

The probabiility of losing this bet is the sum of:

- The probability of having a face card in the first turn

- The probability of having a face card in the second turn, having a non-face card in the first turn.

- The probability of having a face card in the third turn, having a non-face card in the previous turns.

- The probability of having a face card in the fourth turn, having a non-face card in the previous turns.

<u><em>1) The probability of having a face card in the first turn</em></u>

In this case, the chances are 12 in 52:

$P_1=P(face\, card)=12/52=0.231$

<u><em>2) The probability of having a face card in the second turn, having a non-face card in the first turn.</em></u>

In this case, first we have to get a non-face card (there are 40 in the dech of 52), and then, with the rest of the cards (there are 51 left now), getting a face card:

$P_2=P(non\,face\,card)*P(face\,card)=(40/52)*(12/51)=0.769*0.235=0.181$

<u><em>3) The probability of having a face card in the third turn, having a non-face card in the first and second turn.</em></u>

In this case, first we have to get two consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_3=(40/52)*(39/51)*(12/50)\\\\P_3=0.769*0.765*0.240=0.141$

<u><em>4) The probability of having a face card in the fourth turn, having a non-face card in the previous turns.</em></u>

In this case, first we have to get three consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_4=(40/52)*(39/51)*(38/50)*(12/49)\\\\P_4=0.769*0.765*0.76*0.245=0.109$

With these four probabilities we can calculate the probability of losing this bet:

$P=P_1+P_2+P_3+P_4=0.231+0.181+0.141+0.109=0.662$

The probability of losing is 66.2%, which is the same as saying you have (1-0.662)=0.338 or 33.8% of winning chances. Losing is more probable than winning, so you should not take the bet.

b) If the bet involves 3 cards, the only difference with a) is that there is no probability of getting the face card in the fourth turn.

We can calculate the probability of losing as the sum of the first probabilities already calculated:

$P=P_1+P_2+P_3=0.231+0.181+0.141=0.553$

There is 55.3% of losing (or 44.7% of winning), so it is still not convenient to bet.

5 0

4 years ago

Given the two points what is the slope of the line?<br> (4,2) (10.4)

zlopas [31]

$m = \frac{y2-y1}{x2-x1}\\ m=\frac{4-2}{10-4} \\m = \frac{2}{6} \\\\m=\frac{1}{3}$

Therefore, the slope of the line is 1/3. In case if you want the equation too.

$y=mx+b\\y=\frac{1}{3}x+b\\ 2=\frac{1}{3}(4)+b\\ 2=\frac{4}{3}+b\\ b=-\frac{4}{3}+2\\ b=-\frac{4}{3} +\frac{6}{3}\\ b=\frac{2}{3}$

Therefore, the equation is y=1/3x+2/3

6 0

3 years ago

What is the value of x3 × y1 when x = 5 and y = 10

katovenus [111]

Answer:

150

Step-by-step explanation:

To solve the problem, put 5 instead of x, so 5*3

and put 10 instead of y, so 10*1

when combined,

(5*3) * (10*1)

= 15 * 10

= 150

Hope this helps!!

Let me know if I'm wrong...

8 0

3 years ago