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Ivanshal [37]
2 years ago
7

Suppose you are researching the eating habits of people your age. What sampling method could you use to find the percentage of s

tudents in your grade who eat five servings of fruit and vegetables each day? What is an example of a survey question that does not have bias?
Mathematics
2 answers:
11Alexandr11 [23.1K]2 years ago
8 0
Your best decision would be to find a subject that the entire grade has so it is not biased such as a course class like science or language arts depending on what grade they are referring to if this is not the answer you where looking for I am sorry hope I helped :)
sveta [45]2 years ago
6 0

Answer: Asking all people in grade, and i would ask "how many servings of vegetables and fruits do you generally eat a day?

Step-by-step explanation:I wouldn't stick to asking only say the wrestling team do to high carb diets but i also wouldn't just ask your friends either  because some of you couldn't all have the money to afford fruits and vegetables since those are generally higher prices

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Wilson has a balance of $890 on a credit card with an apr of 18.7%, compounded monthly. about how much will he save in interest
vivado [14]
890×(1+0.187÷12)^(12)−890×(1+0.125÷12)^(12)=63.61....answer
5 0
2 years ago
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Kelly purchased 6 planters for a total of $18. She wants to purchase another 16 planters at the same unit price. How much will 1
AfilCa [17]

Answer:

Hope this helps

Step-by-step explanation:

If 6 planters= $18   then 16 planters= 16x  18 divided by 6 =$48

5 0
2 years ago
At a particular restaurant, each onion ring has 45 calories and each slider has 325 calories. A combination meal with onion ring
AlladinOne [14]

Answer:

There were 6 onion rings and 2 slider in the combination meal.

Step-by-step explanation:

Let x be the number of onion rings and y be the number of slider.

Calories in one onion ring = 45 calories

Calories in one slider = 325 calories

Total number of calories = 920

Thus, we can write the equation:

45x + 325y = 920

There are 3 times as many onion rings as the sliders.

Thus, we can write the equation:

x = 3y

Solving the two equation by substitution method, we get,

45(3y) + 325y = 920\\460y = 920\\y = 2\\x = 3y = 3(2) = 6

Thus, there were 6 onion rings and 2 slider in the combination meal.

7 0
2 years ago
Find a set of scalar parametric equations for the line formed by the two intersecting planes. 3x+3y+2z+2=0 2x−3y+2z−2=0
vivado [14]

The normal vectors to the two planes are (3, 3, 2) and (2, -3, 2). The cross product of these will be the direction vector of the line of intersection, (12, -2, -15).

Using x=0, we can find a point on this line by solving the simultaneous equations that remain:

... 3y +2z = -2

... -3y +2z = 2

Adding these, we get

... 4z = 0

... z = 0

so the point we're looking for is (x, y, z) = (0, -2/3, 0). This gives rise to the parametric equations ...

  • x = 12t
  • y = -2/3 -2t
  • z = -15t

By letting t=2/3, we can find a point on the line that has integer coefficients. That will be (x, y, z) = (8, -2, -10).

Then our parametric equations can be written as

  • x = 8 +12t
  • y = -2 -2t
  • z = -10 -15t
8 0
2 years ago
Uranus has a mass of 8.68 1025 kg and a radius of 2.56 107 m. Assume it is a uniform solid sphere. The distance of Uranus from t
Lelechka [254]

Answer

Given,

Mass of the Uranus, M = 8.68 x 10²⁵ Kg

Radius of Uranus, R = 2.56 x 10⁷ m

Distance of Uranus, D = 2.87 x 10¹² days

a) Rotational Kinetic energy of the Uranus

moment of inertia of the Uranus

  I = \dfrac{2}{5}MR^2

  I = \dfrac{2}{5}\times 8.68\times 10^{25}\times (2.56\times 10^7)^2

         I = 22.75 x 10³⁹ kg.m²

  Angular speed

     \omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{17.3\times 3600}\

    \omega = 1 \times 10^{-4}

 Rotational Kinetic energy

     KE = \dfrac{1}{2}I\omega^2

     KE = \dfrac{1}{2}\times 22.75\times 10^{39}\times (10^{-4})^2

      KE = 11.38\times 10^{31}\ J

b) Rotational Kinetic energy of Uranus in its orbit around sun

moment of inertia of the Uranus

  I = \dfrac{2}{5}MR^2+ Ma^2

  I = 22.75\times 10^{39}+ 8.68\times 10^{25}\times (2.87\times 10^{12})^2

      I = 7.15 x 10⁵⁰ kg.m²

Angular speed

     \omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{3.08\times 10^4\times 3600\times 24}\

    \omega =2.36\times 10^{-9}

 Rotational Kinetic energy

     KE = \dfrac{1}{2}I\omega^2

     KE = \dfrac{1}{2}\times 7.15\times 10^{50}\times (2.36\times 10^{-9})^2

      KE = 1.99\times 10^{33}\ J

6 0
2 years ago
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