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MAVERICK [17]
2 years ago
7

Sharon and Jacob started at the same place. Jacob walked 3 m north and then 4 m west. Sharon walked 5 m south and 12 m east. How

far apart are Jacob and Sharon now?
Mathematics
1 answer:
Ilya [14]2 years ago
7 0

Consider the coordinate plane:

1. The origin is the point where Sharon and Jacob started - (0,0).

2. North - positive y-direction, south - negetive y-direction.

3. East - positive x-direction, west - negative x-direction.

Then,

  • if Jacob walked 3 m north and then 4 m west, the point where he is now has coordinates (-4,3);
  • if Sharon walked 5 m south and 12 m east, the point where she is now has coordinates (12,-5).

The distance between two points with coordinates (x_1,y_1) and (x_2,y_2) can be calculated using formula

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.

Therefore, the distance between  Jacob and Sharon is

D=\sqrt{(12-(-4))^2+(-5-3)^2}=\sqrt{16^2+8^2}=\sqrt{256+64}=\sqrt{320}=8\sqrt{5}\approx 11.18\ m.

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Answer:

D. The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

Step-by-step explanation:

The treatment group using System A reported a mean of 18.5 lost bags per day. The treatment group using System B reported a mean of 16.6 lost bags per day.

The best conclusion that can be made is - The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

As we know, in statistics, nothing happens by chance. So, this option is correct.

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2 years ago
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Line RS intersects triangle BCD at two points and is parallel to segment DC. Triangle B C D is cut by line R S. Line R S goes th
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Answer:

The correct options are;

1) ΔBCD is similar to ΔBSR

2) BR/RD = BS/SC

3) (BR)(SC) = (RD)(BS)

Step-by-step explanation:

1) Given that RS is parallel to DC, we have;

∠BDC = ∠BRS (Angles on the same side of transversal)

Similarly;

∠BCD = ∠BSR (Angles on the same side of transversal)

∠CBD = ∠CBD = (Reflexive property)

Therefore;

ΔBCD ~ ΔBSR Angle, Angle Angle (AAA) rule of congruency

2) Whereby  ΔBCD ~ ΔBSR, we therefore have;

BC/BS = BD/BR → (BS + SC)/BS = (BR + RD)/BR = 1 + SC/BS = RD/BR + 1

1 + SC/BS = 1 + RD/BR = SC/BS = 1 + BR/RD - 1

SC/BS = RD/BR

Inverting both sides

BR/RD = BS/SC

3) From BR/RD = BS/SC the above we have by cross multiplication;

BR/RD = BS/SC gives;

BR × SC = RD × BR → (BR)(SC) = (RD)(BR).

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2 years ago
The sales data for January and February of a frozen yogurt shop are approximately normal. The mean daily sales for January was $
VashaNatasha [74]

Answer:

January had a higher z-score for sales on the 15th, and the value of that z-score was of 0.5.

Step-by-step explanation:

z-score:

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

January:

The mean daily sales for January was $300 with a standard deviation of $20. On the 15th of January, the shop sold $310 of yogurt. This means, respectively, that \mu = 300, \sigma = 20, X = 310. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{310 - 300}{20}

Z = 0.5

February:

The mean daily sales for February was $320 with a standard deviation of $50. On the 15th of February, the shop sold $340 of yogurt. This means, respectively, that \mu = 320, \sigma = 50, X = 340. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{340 - 320}{50}

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January had a higher z-score for sales on the 15th, and the value of that z-score was of 0.5.

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In a lecture demonstration, an object is suspended from a spring scale which reads 8N when the object is in air. The object is t
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Answer:

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volume of water displaced = 0.204/1000 = 0.000204m^3 (204cm^3)

volume of water displaced = volume of the solid

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b) when fully submerge in water the the scale experience according to newton third law of motion ( equal and opposite reaction of forces) additional 2N push so that total weight with the fully submerge solid = 20N + upthrust = 20N + 2N =22N

c) the of two scale reading is before (8N + 20N = 28N) and after (6N + 22N = 28) since there is no loss of matter; the demonstration was in equilibrium.

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