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Lelechka [254]
3 years ago
7

The weights of red delicious apples are approximately normally distributed with a mean of 9 Ounces and a standard deviation of 0

.75 ounce. An online gift store sells gift boxes containing delicious apples. At the time of packaging. 5 red delicious apples are randomly selected and packaged in a box. red a. Describe the distribution of the total weight of the 5 randomly selected apples. b. What is the probability that the total weight of the 5 randomly selected apples will be less than 42 ounces? c. The combined weight of the packing material and box in which the apples will be shipped is always 10 ounces. Let W represent the weight of a complete packaged gift box, which consists of the packing material, box, and 5 randomly selected apples. What are the mean and the standard deviation of W?
Mathematics
1 answer:
Simora [160]3 years ago
3 0

Answer:

The mean of W is 55 ounces.

The standard deviation of W is 4.33 ounces.

Step-by-step explanation:

Let X: weight of a red delicious apple, and B: the weight of the box and packing material.

The distribution that will represent W: the total weight of the packaged 5 randomly selected apples will be also normally distributed.

Applying the property of the mean: E(aX)=aE(X), the mean of W will be:

\mu_W=E(W)=E(5X+B)=5E(X)+E(B)=5*9+10=45+10=55

Applying the property of the variance: V(aX+b)=a^2V(X), the variance of W will be:

\sigma^2_W=V(W)=V(5X+B)=5^2V(X)+V(B)=25*0.75+0=18.75

The mean standard deviation of W will be the squared root of V(W):

\sigma_W=\sqrt{V(W)}=\sqrt{18.75}=4.33

The mean of W is 55 ounces.

The standard deviation of W is 4.33 ounces.

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Answer:

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The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg
julia-pushkina [17]

Answer:

Approximately 101\; \rm ft (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

  • Let \rm O denote the observer.
  • Let \rm A denote the top of the tree.
  • Let \rm R denote the base of the tree.
  • Let \rm B denote the point where line \rm AR (a vertical line) and the horizontal line going through \rm O meets. \angle \rm B\hat{A}R = 90^\circ.

Angles:

  • Angle of elevation of the base of the tree as it appears to the observer: \angle \rm B\hat{O}R = 51^\circ.
  • Angle of elevation of the top of the tree as it appears to the observer: \angle \rm B\hat{O}A = 57^\circ.

Let the length of segment \rm BR (vertical distance between the base of the tree and the base of the hill) be x\; \rm ft.

The question is asking for the length of segment \rm AB. Notice that the length of this segment is \mathrm{AB} = (x + 20)\; \rm ft.

The length of segment \rm OB could be represented in two ways:

  • In right triangle \rm \triangle OBR as the side adjacent to \angle \rm B\hat{O}R = 51^\circ.
  • In right triangle \rm \triangle OBA as the side adjacent to \angle \rm B\hat{O}A = 57^\circ.

For example, in right triangle \rm \triangle OBR, the length of the side opposite to \angle \rm B\hat{O}R = 51^\circ is segment \rm BR. The length of that segment is x\; \rm ft.

\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}.

Rearrange to find an expression for the length of \rm OB (in \rm ft) in terms of x:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}.

Similarly, in right triangle \rm \triangle OBA:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}.

Equate the right-hand side of these two equations:

0.810\, x \approx 0.649\, (x + 20).

Solve for x:

x \approx 81\; \rm ft.

Hence, the height of the top of this tree relative to the base of the hill would be (x + 20)\; {\rm ft}\approx 101\; \rm ft.

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Answer:

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423.9/ 3.14 = 135  so 135 is the diameter.

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Now is time to find the area which uses the formula  A= nr^2

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