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3 years ago

To the nearest square unit, what is the area of the regular heptagon shown below?

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

4 0

Answer:

B. 1783 square units

Step-by-step explanation:

user100 [1]3 years ago

3 0

Answer:

Its b: 1572 square units

Step-by-step explanation:

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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 X represent the amount of money that the family has invested in different real estate properties.

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

It is provided that the family has invested in n = 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 z is 0.84.

*Use a z-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

3 years ago

Kyla spends 60 Minutes of each day exercising. Let d be the number of days that Kyla exercises, and let m represent the total nu

stealth61 [152]

Answer: There is linear relationship between the number of days that Kyla exercise in the total minutes that she exercises.

The independent variable is 'd' and m is the dependent variable which depends on the number of days she exercise.

The linear equation for the situation is given by

$m=60d$

Step-by-step explanation:

Let d be the number of days that Kyla exercises, and let m represent the total numbers of minutes she exercise.

Kyla spends 60 Minutes of each day exercising which is constant .

Then the total numbers of minutes she exercise(m) in d days is given by

$m=60d$ which is the linear equation.

The relationship between the number of days that Kyla exercise in the total minutes that she exercises is linear, where d is the independent variable, and m is the dependent variable which depends on the number of days she exercise.

[ad d increases m increases by rate of 60 minutes per day]

The linear equation for the situation is given by

$m=60d$

7 0

3 years ago

Find the total area the regular pyramid. T. A. =

lorasvet [3.4K]

Answer:

T.A. = 16√3 units²

Step-by-step explanation:

∵ The total area = the area of the four faces

∵ The four faces are equilateral triangles with side length 4

∵ Area of the equilateral Δ = 1/4 s² √3

∴ T.A. = 4 × 1/4 × 4² × √3 = 16√3 units²

5 0

3 years ago

Read 2 more answers

Translate 5x+7 into a verbal sentence.

algol13

B im pretty sure....

7 0

2 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago