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Rudik [331]
2 years ago
9

Janelle has 4 hours to spend training for an upcoming race. She completes her training by running full speed the distance of the

race and walking back the same distance to cool down. If she runs at a speed of 7mph and walks back at a speed of 3mph how long should she plan to spend walking back?
Mathematics
1 answer:
natima [27]2 years ago
8 0
If the amount of time taken to go said distance is x and the amount of time taken to go back said distance is y, then the amount of miles total is 7x+3y due to that for every hour, she adds 7 miles when going there and 3 miles for walking back. In addition, since the total amount of time is 4 hours, x+y=4 as the total time spent as well as 7x=3y due to that they're the same distance. 
x+y=4
7x=3y
Dividing the second equation by 7, we get x=3y/7. Plugging that into the first equation, we get 3y/7+y=4=10y/7 (since y=7y/7). Multiplying both sides by 7 and then dividing both by 10, we get 28/10=2.8=y in hours. Since 0.1 hours is 60/10=6 minutes, and 0.8/0.1=8, 6*8=48 minutes=0.8 hours, meaning that she should plan to spend 2 hours and 48 minutes walking back 

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A village experienced 2% population growth, compounded continuously, each year for 10 years. At the end of the 10 years, the pop
pantera1 [17]

Answer:

The initial population at the beginning of the 10 years was 129.

Step-by-step explanation:

The population of the village may be modeled by the following function.

P(t) = P_{0}e^{rt}

In which P is the population after t hours, P_{0} is the initial population and r is the growth rate, in decimal.

In this problem, we have that:

P(10) = 158, r = 0.02.

So

158 = P_{0}e^{0.02*10}

P_{0} = 158*e^{-0.2}

P_{0} = 129

The initial population at the beginning of the 10 years was 129.

6 0
3 years ago
Read 2 more answers
2. A teacher asked his class of 20 students, "What is your age? Their responses are shown in the dot plot below.
Karolina [17]

Answer:

Yes, average does represent the centre of this data.

Average = 15.9

Step-by-step explanation:

As this question is not complete and not correctly written. Let's try to find an answer for this question.

Total number of Students = 20 = n

and we are given this data:

14, 15, 17, 18, 16.

which is just five entries and total number of students are 20. 15 entries are missing in this questions. So, let's add our own values into this question.

New set of entries for 20 students.

1. 14

2. 15

3. 17

4. 18

5. 16

6. 19

7. 17

8. 16

9. 17

10. 17

11. 15

12. 20

13. 14

14. 17

15. 18

16. 19

17. 13

18. 20

19. 17

20. 16

So, let's calculate the average of these numbers.

Average = Sum of all entries/ Total number of entries.

Average = 318/20

Average = 15.9

So, the average age of all the students of class is 15.9.

Let's calculate median:

For median , data must be arranged in a order from least to greatest number. Median is a number from that data which is at half way. In case of odd numbers, it is easy to identify median because there is only one number. and in case of even numbers, median is the average of the two middle numbers. So, here we have even numbers and median is the average of two middle numbers = 17 + 17/2 = 17

Median = 17

Yes, average does represent the centre of this data.

8 0
2 years ago
Read 2 more answers
What is the square root of
Veronika [31]
Sq rt of 8464= 92
sq rt of 29584= 172
sq rt of 15376= 124
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sq rt of 44100= 210
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5 0
2 years ago
For a certain population of men, 8 percent carry a certain genetic trait. For a certain population of women, 0.5 percent carry t
slavikrds [6]

Answer:

C) 150 men and 100 women

D) 200 men and 2000 women

E) 1000 men and 1000 women

Step-by-step explanation:

Hello!

To compare the proportion of people that carry certain genetic trait in men and woman from a certain population two variables of study where determined:

X₁: Number of men that carry the genetic trait.

X₁~Bi(n₁;p₁)

X₂: Number of women that carry the genetic trait.

X₂~Bi(n₂;p₂)

The parameter of interest is the difference between the population proportion of men that carry the genetic trait and the population proportion of women that carry the genetic trait, symbolically: p₁-p₂

To be able to study the difference between the population proportions you have to apply the Central Limit Theorem to approximate the distribution of both sample proportions to normal.

<u><em>Reminder:</em></u>

Be a variable with binomial distribution X~Bi(n;p), if a sample of size n is taken from the population in the study. Then the distribution of the sample proportion tends to the normal distribution with mean p and variance (pq)/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

So for both populations in the study, the sample sizes should be

n₁ ≥ 30

n₂ ≥ 30

Also:

Both samples should be independent and include at least 10 successes and 10 failures.

Both populations should be at least 20 times bigger than the samples. (This last condition is to be assumed because without prior information about the populations is impossible to verify)

  • If everything checks out then (p'₁-p'₂)≈N(p₁-p₂; p(1/n₁+1/n₂))

<u>The options are:</u>

A) 30 men and 30 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 30*0.08= 2.4

Failures: y₁= n₁*q₁= 30*0.92= 27.6

Population 2

Successes: x₂= n₂*p₂= 30*0.5= 15

Failures: y₂= n₂*q₂= 30*0.5= 15

The second condition is not met.

B) 125 men and 20 women

n₁ ≥ 30 but n₂ < 30

Both samples are independent but n₂ is not big enough for the approximation.

C) 150 men and 100 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 150*0.08= 12

Failures: y₁= n₁*q₁= 150*0.92= 138

Population 2

Successes: x₂= n₂*p₂= 100*0.5= 50

Failures: y₂= n₂*q₂= 100*0.5= 50

All conditions are met, an approximation to normal is valid.

D) 200 men and 2000 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 200*0.08= 16

Failures: y₁= n₁*q₁= 200*0.92= 184

Population 2

Successes: x₂= n₂*p₂= 2000*0.5= 1000

Failures: y₂= n₂*q₂= 2000*0.5= 1000

All conditions are met, an approximation to normal is valid.

E) 1000 men and 1000 women

n₁ ≥ 30 and n₂ ≥ 30

Both samples are big enough and independent.

Population 1

Successes: x₁= n₁*p₁= 1000*0.08= 80

Failures: y₁= n₁*q₁= 1000*0.92= 920

Population 2

Successes: x₂= n₂*p₂= 1000*0.5= 500

Failures: y₂= n₂*q₂= 1000*0.5= 500

All conditions are met, an approximation to normal is valid.

I hope this helps!

6 0
3 years ago
Wright et al. [A-2] used the 1999-2000 National Health and Nutrition Examination Survey(NHANES) to estimate dietary intake of 10
xxMikexx [17]

Complete question :

Wright et al. [A-2] used the 1999-2000 National Health and Nutrition Examination Survey NHANES) to estimate dietary intake of 10 key nutrients. One of those nutrients was calcium in all adults 60 years or older a mean daily calcium intake of 721 mg with a standard deviation of 454. Usin these values for the mean and standard deviation for the U.S. population, find the probability that a randonm sample of size 50 will have a mean: (mg). They found a) Greater than 800 mg b) Less than 700 mg. c) Between 700 and 850 mg.

Answer:

0.10935

0.3718

0.9778

0.606

Step-by-step explanation:

μ = 721 ; σ = 454 ; n = 50

P(x > 800)

Zscore = (x - μ) / σ/sqrt(n)

P(x > 800) = (800 - 721) ÷ 454/sqrt(50)

P(x > 800) = 79 / 64.205295

P(x > 800) = 1.23

P(Z > 1.23) = 0.10935

2.)

Less than 700

P(x < 700) = (700 - 721) ÷ 454/sqrt(50)

P(x < 700) = - 21/ 64.205295

P(x < 700) = - 0.327

P(Z < - 0.327) = 0.3718

Between 700 and 850

P(x < 850) = (850 - 721) ÷ 454/sqrt(50)

P(x < 850) = 129/ 64.205295

P(x < 700) = 2.01

P(Z < 2.01) = 0.9778

P(x < 850) - P(x < 700) =

P(Z < 2.01) - P(Z < - 0.327)

0.9778 - 0.3718

= 0.606

3 0
2 years ago
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