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Find the area of the rhombus. d1 = 12 m; d2 = 20 m

kati45 [8]

Check the picture below.

so, assuming d1 and d2 are the diagonals of it, bearing in mind that the diagonals of a rhombus bisect each, cut in equal halves, then we can get the area of one of those 4 congruent triangles in the rhombus, notice each triangle has a base of 6 and a height of 10.

$\bf \stackrel{\textit{area of the 4 triangles}}{4\left[ \cfrac{1}{2}(6)(10) \right]}$

5 0

3 years ago

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The weights of a certain brand of candies are normally distributed with a mean weight of 0.8556 g and a standard deviation of 0.

babunello [35]

Answer:

a) There is a probability P=0.4992 that a randomly selected candy weights at least 0.8543 g.

b) There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c) The claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

Step-by-step explanation:

The average weight of the candies in the package is:

$\bar x=\dfrac{\sum x_i}{n}=\dfrac{400.3}{469}=0.8535$

For a randomly selected candy (is a sample of size n=1) the standard deviation is the population standard deviation (sigma=0.0511 g).

The probabiltiy that is weights more than 0.8536 can be calculated with the z-score.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}= \dfrac{0.8536-0.8535}{0.0511/\sqrt{1}}=\dfrac{0.0001}{0.0511}=0.002$

$P(X>0.8536)=P(z>0.002)=0.4992$

There is a probability P=0.4992 that a randomly selected candy weights at least 0.8536 g.

b. If the sample is now of n=441 candies, and we want to know the probability that the mean weight is at least 0.8543 g, the z-score needs to be recalculated:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8543-0.8535}{0.0511/\sqrt{441}}=\dfrac{0.0008}{0.0024}=0.3288$

$P(X_s>0.8543)=P(z>0.3288)=0.37115$

There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c. To be more confident about the claim that the mean weight is 0.8556 g, we can calculate the probability that, for a package of 469 candies and using the mean of 0.8535 g that we calculated before, the average weight is at least 0.8556 g.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8556-0.8535}{0.0511/\sqrt{469}}=\dfrac{0.0021}{0.0024}=0.89$

$P(X_s>0.8556)=P(z>0.89)=0.18673$

The probability is P=0.187, so the claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

6 0

3 years ago

You work at a pioneer historical site. On this site you have handcarts. One cart has a handle that connects to the center of the

erastovalidia [21]

Let the center of the wheel be O, and let the endpoint of the handle as shown in the original picture be H.

Check the first picture attached. OH=48 and m(OHB)=20°.

|OB|=|OH|*sin20°=48 in * 0.342= 16.4 in

The handle of the handcart when lifted, moves around a circle formed, with center and radius |OH|=48 in.

To elevate the handle at 48 in off the ground, the handle first comes at point H', such that OH'//BH, then comes to point H" such that the distance of point H" to the ground is 48 inches. (check picture 1)

check picture 2, the distance of the point H'' to the line segment OH' is 48-16.4= 31.6 (inches)

31.6 = 48 * sinα

sin α=31.6/48=0.6583

so α = arcsin(0.6583)=41.17°

similarly, instead of α, 20° is the greatest angle such that the content dont spill out, and let h be the height above the line segment OH'.

h= 48*sin20=48*0.342=16.4

so the maximal height at which the content dont spill is 16.4+16.4=32.8 (inches)

Answer: 1. the contents would spill out, 2. maximal height: 32.8 inches