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Sergio039 [100]
3 years ago
14

The jogging track is of a mile long. If Ashley jogged around it 4 times, how far did she run? A. B. C. D.

Mathematics
2 answers:
OlgaM077 [116]3 years ago
8 0

Answer:

she did 4 miles have a nice day

Step-by-step explanation:

Temka [501]3 years ago
7 0

Answer:

4.00000000 miles

Step-by-step explanation:

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Help me please and thank you.
dezoksy [38]

Answer:

0.67x-2=0.43

Step-by-step explanation:

2/3 = 0.6 repeating

3/7 = 0.42857

7 0
3 years ago
last year, 639 students attended a summer camp. of those who attended this year, 0.5% also attended summer camp last year. about
lawyer [7]

Answer:

First of all, he have to find out what 0.5% of the students is.

Second of all, find out what 0.5 of 100 is. 100 divided by 0.5 is 200.

Third of all, divide 200 from 639. This equals 3.195

Since it is asking about how many, I would go with about 3. After all, you can't have have half of a person, so the decimals shouldn't be added.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
QUESTION
Lorico [155]
1. For this item we just refer to the prompt to know the conjectures of Ernest and Denise. According to Ernest, they should swim 1 kilometer on the first week then add 0.25km every week while Denise believes that they should swim 1 kilometer on the first week then add 0.5km every week.

2. Yes, these distances make an arithmetic sequence. It's because an arithmetic sequence is defined as a group of increasing or decreasing numbers where the difference between any two consecutive numbers is constant. This just means that every number has the same interval. In the case of their schedule, this is true.

3. For this item we just follow the descriptions of Ernest's and Denise's schedule in item number 1. For Ernest, we just keep adding 0.25 from 1 kilometer until we added it thrice. For Denise, we also keep adding a number thrice but this time it's 0.5 instead of 0.25.

Ernest's Schedule: 1, 1.25, 1.5, 1.75
Denise's Schedule: 1, 1.5, 2, 2.5

4. Here we are asked to determine a formula that will describe the schedules of Ernest and Denise. In the given formula a_{n}= a_{n-1}+d, a_{n} refers to the next term in the sequence, a_{n-1} refers to the previous term, while d refers to the common difference. In the recursive formula all we need is to insert the value of d to the equations.

Ernest: a_{n}= a_{n-1}+0.25
Denise: a_{n}= a_{n-1}+0.5

5. For this item we basically do the same thing but this time we are given another formula. Our formula is in the form a_{n}= a_{1}+(n-1)d where a_{n} is still the nth term of the sequence, a_{1} is the very first time, n is the number of terms, and d is the common difference. 

Ernest: a_{n}= 1.0+0.25(n-1)
Denise: a_{n}= 1.0+0.5(n-1)

6. In this item we will just basically substitute numbers to one of the equations that we've set up in item #5. For this we need Ernest's explicit formula first. To know how far they will be swimming on week 10, the number of elements (n) must be 10.

a_{10}= 1.0+0.25(10-1)
a_{10}= 1.0+0.25(9)
a_{10}= 1.0+2.25
a_{10}= 3.25

7. Here, we just do the same thing as item #6 but this time we will consider Denise's explicit formula. Since we are also asked how far the students will be swimming on week 10, the number of elements would also be 10 and this would also be our value for n.

a_{10}= 1.0+0.5(10-1)
a_{10}= 1.0+0.5(9)
a_{10}= 1.0+4.5
a_{10}= 5.5

8. The answer for this question is obvious. You would just need to look at the 10th element in Ernest's and Denise's sequences and tell whose schedule had more than or equal to 5 as an answer. Following Ernest's schedule, you will just get 3.5 kilometers on the 10th week so it's definitely a no. Denise's schedule, on the other hand, would get you to 5.5 kilometers on week 10 so her training schedule should be followed.
7 0
3 years ago
A football team has a probability of .75 of winning when playing any of the other four teams in its conference. If the games are
Alexeev081 [22]

Answer:

0.3164 = 31.64% probability the team wins all its conference games

Step-by-step explanation:

For each conference game, there are only two possible outcomes. Either the team wins it, or they lose. The probability of winning a game is independent of any other game. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A football team has a probability of .75 of winning when playing any of the other four teams in its conference.

The probability means that p = 0.75, and four games means that n = 4

If the games are independent, what is the probability the team wins all its conference games?

This is P(X = 4). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{4,4}.(0.75)^{4}.(0.25)^{0} = 0.3164

0.3164 = 31.64% probability the team wins all its conference games

4 0
3 years ago
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