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3 years ago

PLEASE HELP. WILL GIVE BRAINLEST

Mathematics

1 answer:

swat323 years ago

5 0

Answer:

i would say A. The initial cost for renting a snowmobile is $75, with each hour of use costing an additional $25.

Step-by-step explanation:

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Solve for w, simplify your answer as much as possible

Akimi4 [234]

Answer:

W=3

Step-by-step explanation:

(6w + 4w)/48=5/8(taking lcm)

6w+4w=5/8*48

10w=30

W=3

Thanks for putting qsn, makes me remember this againa

6 0

2 years ago

Christov and Mateus gave out candy to children on Halloween. They each have out candy at a constant rate, and they both gave awa

padilas [110]

Answer: 1) Both gave same number of candies to each child.

2) Christov gave candy to a greater number of children.

Step-by-step explanation:

Since, Christov initially had 300 pieces of candy, and after he was visited by 17 children, he had 249 pieces left.

Let he gave x candy to each student when he visited 17 children.

Then, 17 x + 249 = 300

⇒ 17 x = 51

⇒ x = 3

Since, he distributes candies in a constant speed.

Therefore, his speed of distribution = 3 candies per child

Now,The function that shows the remaining number of candies Mateus has after distributing candies to n children,

C(n)=270 - 3 n

Initially, n = 0

C(0) = 270

⇒ Mateus has initial number of candies = 270

When he gave candies to one child then remaining candies = 270 - 3 × 1 = 267

Thus, the candies, get by a child from Mateus = 270 - 267 = 3

Since, he distributes candies in a constant speed.

⇒ His speed of distribution = 3 candies per child

1) Therefore, Both Christov and Mateus have same speed of distribution.

2) Since, both have same seed.

⇒ The one who has greater number of candies will be distribute more.

⇒ Christov will give more candies.

8 0

3 years ago

Do any of you get this question? Please help!

Alexxandr [17]

Honestly i just wanted to try this question since ive never seen it but i dont rlly know if i did it right at all

I equaled GH, HI, and GI together to get y

which i got y=-2

and what i got next i feel is off since

GH, HI, and GI all equaled -11

if you kind of know how to do the question maube you could correct me from there but otherwise dont take my word for it completely

4 0

3 years ago

Help asap pleeeas!

olga_2 [115]

The answer to this question is C.2 I think

6 0

3 years ago

Read 2 more answers

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.

1) The probability of having a face card in the first turn

In this case, the chances are 12 in 52:

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

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

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:

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

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

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:

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

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

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

3 years ago