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tiny-mole [99]
3 years ago
14

PLZZ ANSWER 25 PTS RATED ATAR PROMISE TO GIVE BRAINLIEST PLZZZ ANYONE KNOW THIS???? A 154-lb person burns 420 calories per hour

riding an exercise bicycle at a rate of 15 mi/hr. Write a function rule to represent the total calories burned over time by that person. Explain how the information in the problem relates to the function.
Mathematics
2 answers:
Novosadov [1.4K]3 years ago
8 0

Answer:

c = 420t . . . . c is calories burned; t is hours riding at 15 mph

Step-by-step explanation:

There is not enough information given to write a function rule relating all the variables to calories burned. If we assume that calories are burned at the constant rate of 420 calories per hour, then total calories will be that rate multiplied by hours:

c = 420·t

where c is total calories burned by the 154-lb person, and t is hours riding at 15 mph.

___

In general, rates are related to quantities by ...

quantity = rate · time . . . . . where the rate is (quantity)/(time period)

Sergeu [11.5K]3 years ago
7 0

Answer:

well it would be c=420t but in reality it is just 420t

Step-by-step explanation:

They are only asking for two parts out of three. The 154-lb part is dropped from the equation leaving us with

  • x = (calories)*time which is just the base.
  • C=420t in simple terms is just the total calories over time.

C is calories, T is time; you're left with your starting base which is 420 calories per hour * the amount of time he spent biking.

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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
What is the surface area of the square pyramid represented by the net? Enter your answer in the box. m²
Alborosie

Answer:

1.75 meters² or 175 centimeters²

Step-by-step explanation:

Step 1: Find Area of the Base.

  • Base is a square with dimensions of 7cm by 7cm
  • 7*7
  • 49 cm²

Step 2: Find the Area of the Triangles.

  • Each triangle has a base of 7cm and a height of 9cm.
  • 7 x 9 = 63, and 63/2 = 31.5
  • 31.5 x 4 = 126 cm²

Step 3: Add the areas together.

  • 49+126
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Thus, the surface area is 1.75 m^2 or 175 cm^2.

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Answer:

a. 0.1536

b. 0.9728

Step-by-step explanation:

The probability that a component fails, P(Y) = 0.2

The number of components in the system = 4

The number of components required for the subsystem to operate = 2

a. By binomial theorem, we have;

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536

b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working

Therefore, we have;

P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)

P(Y = 0) = \dbinom{4}{0} × 0.2⁰ × 0.8⁴ = 0.4096

P(Y = 1) = \dbinom{4}{1} × 0.2¹ × 0.8³ = 0.4096

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728

The probability that the subsystem operates longer than 1,000 hours = 0.9728

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Answer:

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D

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