A soccer field has an area of 7/8 square miles. The length of the field is 2/5 miles. What is the width of the soccer field?

What is likely to have a mass of 1 gram

Gre4nikov [31]

Answer:

paperclip, thumbtack, or safety pin.

Step-by-step explanation:

I measured some.

4 0

2 years ago

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A financial advisor is analyzing a family's estate plan. The amount of money that the family has invested in different real esta

Pachacha [2.7K]

Answer:

The amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

Step-by-step explanation:

Let the random variable <em>X</em> represent the amount of money that the family has invested in different real estate properties.

The random variable <em>X</em> follows a Normal distribution with parameters <em>μ</em> = $225,000 and <em>σ</em> = $50,000.

It is provided that the family has invested in <em>n</em> = 10 different real estate properties.

Then the mean and standard deviation of amount of money that the family has invested in these 10 different real estate properties is:

$\mu_{\bar x}=\mu=\$225,000\\\\\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{50000}{\sqrt{10}}=15811.39$

Now the lowest 80% of the amount invested can be represented as follows:

$P(\bar X$

The value of <em>z</em> is 0.84.

*Use a <em>z</em>-table.

Compute the value of the mean amount invested as follows:

$\bar x=\mu_{\bar x}+z\cdot \sigma_{\bar x}$

$=225000+(0.84\times 15811.39)\\\\=225000+13281.5676\\\\=238281.5676\\\\\approx 238281.57$

Thus, the amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

6 0

2 years ago

Help me with this question

TEA [102]

Answer:

the second one

Step-by-step explanation:

5 0

2 years ago

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If a rectangular poster has an area of

d1i1m1o1n [39]

$2x ^2 - 9x + 10=\\
2x^2-4x-5x+10=\\
2x(x-2)-5(x-2)=\\
(2x-5)(x-2) \Rightarrow C$

8 0

2 years ago

You have D dollars to buy fence to enclose a rectangular plot of land (see figure at right). The fence for the top and bottom co

alex41 [277]

The perimeter of the rectangular plot of land is given by the expression below

$P=2x+2y$

On the other hand, since the available money to buy fence is D dollars,

$\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}$

Furthermore, the area of the enclosed land is given by

$A=xy$

Solving the second equation for x,

$\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}$

Substituting into the equation for the area,

$\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}$

To find the maximum possible area, solve A'(y)=0, as shown below

$\begin{gathered} A^{\prime}(y)=0 \\ \Rightarrow\frac{D}{8}-\frac{3}{2}y=0 \\ \Rightarrow\frac{3}{2}y=\frac{D}{8} \\ \Rightarrow y=\frac{D}{12} \end{gathered}$

Therefore, the corresponding value of x is

$\begin{gathered} y=\frac{D}{12} \\ \Rightarrow x=\frac{D-6(\frac{D}{12})}{8}=\frac{D-\frac{D}{2}}{8}=\frac{D}{16} \end{gathered}$ <h2>Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.</h2><h2>As for the used money,</h2> $\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}$ <h2>Half the money was used for the top and the bottom, while the other half was used for the sides.</h2>

7 0

5 months ago