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Charra [1.4K]
3 years ago
6

trans mom said he could spend no more than $12 for rides at the carnival. If the rides cost $0.75 each how many rides can you go

on?
Mathematics
2 answers:
Mariulka [41]3 years ago
7 0

Answer:

16 rides

Step-by-step explanation:

.75 can go into 16, 25 times

kati45 [8]3 years ago
3 0
He can go on no more than 16 rides
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What is the answer? I find this very confusing for me.
vichka [17]

Answer:

19

Step-by-step explanation:

If you want to find the answer of the equation, sub the given information into it:

z + xz + y

we know that:

x = 5

y = 1

z = 3

We got:

= 3 + 5*3 + 1

= 3 + 15 + 1

= 19

Hope this helped :3

5 0
3 years ago
What is the quadratic equation that has roots of ,-3 And 2/5​
prohojiy [21]

Answer: 5x² + 13x - 6 = 0

Step-by-step explanation:

Represent the roots with X i.e.

X = -3 ; X = 2/5

Moving the values of X over the equator we have;

X + 3 = 0 ; 5X - 2 = 0

Equating both equations to zero we can obtain;

(X + 3)(5X - 2) = 0

Opening the brackets we get;

5x² - 2x + 15x - 6 = 0

The  result of the above equation gives;

5x² + 13x - 6 = 0

6 0
3 years ago
What Are The Answers? I Will give Brainliest Answer. (ASAP)
horrorfan [7]

Answer:

1. $8.55

2. $4.29

3. $17.00

4. $10.47

5. $40.31

Step-by-step explanation:

4 0
3 years ago
Sample Response: The two conditional probabilities are not equal because each has different given events. P(A|D) has event D as
Ivan

The conditional probability illustrates that's there's a 2/8 that the event A occurs.

<h3>How to illustrate the probability?</h3>

It should be noted that probability simply means the likelihood of the occurence of an event.

In this case, it can be delivered that P(AID) and P(DIA) aren't equal.

Hence, P(D|A) has event A as its given event, resulting in 2/8 for a probability.

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

7 0
1 year ago
When a fair coin is flipped, what is the chance of getting about 50% heads- specifically between 40% and 60% heads for n=10 flip
aleksklad [387]
At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
3 0
2 years ago
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