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3 years ago

Rudy has been paying an annual homeowners insurance premium of $1,106.30 ($0.37 per $100 of value) since

Mathematics

1 answer:

Luden [163]3 years ago

5 0

Answer:

c. $1,291.30

Step-by-step explanation:

Calculation what will his new annual homeowners insurance premium be

Insurance premium= [$1,106.30+ ($0.37/ $100*$50,000)]

Insurance premium=$1,106.30+$185

Insurance premium=$1,291.30

Therefore what his new annual homeowners insurance premium will be is $1,291.30

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A woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard. The inj

ivanzaharov [21]

Answer:

The maximum possible amount that could be awarded under the two-deviation rule is $ 1.883 Million.

Step-by-step explanation:

In order to find the maximum possible amount of the award, the formula is given as

$x=\bar{x}+2s$

Here

$\bar{x}$ is the mean of the samples and is given as

$\bar{x}= \frac{\sum_{i=1}^{n} x_i}{n}$

s is the standard deviation which is given as

$s=\sqrt{\frac{1}{n-1} (\sum x_i^2-\frac{1}{n} (\sum x_i)^2 )}$

Putting values as

$\sum x_i^2=23,465,528 \$ (1000s) \\ \sum x_i=20,180 \$ (1000s) \\ n=27$

Substituting the values in the equation of mean is as

$\bar{x}= \frac{20,180}{27} \\ \bar{x}= 747.407 \$ (1000s)$

Where as standard deviation is given as

$s=\sqrt{\frac{1}{27-1} (23,465,528-\frac{1}{27} (20,180)^2 )} \\ s=\sqrt{\frac{1}{26} (23,465,528-15082681.4815 )} \\s=\sqrt{\frac{1}{26} (8382846.51)} \\s=\sqrt{322417.17} \\s=567.81 \$ (1000s)$

Putting this in the equation of maximum amount gives

$x=\bar{x}+2.s \\ x=747.407 +2(567.81) \\x=1883.04 \$ (1000s)$

So the maximum possible amount that can be awarded in this case is 1883.04 thousand dollars or $ 1.883 Million. This value is less than the previously maximum value which is $3.5 million.

7 0

3 years ago

Find the fourth roots of 16(cos 200° + i sin 200°).

NeTakaya

Answer:

See below.

Step-by-step explanation:

To find roots of an equation, we use this formula:

$z^{\frac{1}{n}}=r^{\frac{1}{n}}(cos(\frac{\theta}{n}+\frac{2k\pi}{n} )+\mathfrak{i}(sin(\frac{\theta}{n}+\frac{2k\pi}{n})),$ where k = 0, 1, 2, 3... (n = root; equal to n - 1; dependent on the amount of roots needed - 0 is included).

In this case, n = 4.

Therefore, we adjust the polar equation we are given and modify it to be solved for the roots.

Part 2: Solving for root #1

To solve for root #1, make k = 0 and substitute all values into the equation. On the second step, convert the measure in degrees to the measure in radians by multiplying the degrees measurement by $\frac{\pi}{180}$ and simplify.

$z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(0)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(0)\pi}{4}))$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))$

$z^{\frac{1}{4}} = 2(sin(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))$

Root #1:

$\large\boxed{z^\frac{1}{4}=2(cos(\frac{19\pi}{36}))+\mathfrack{i}(sin(\frac{19\pi}{38}))}$

Part 3: Solving for root #2

To solve for root #2, follow the same simplifying steps above but change k to k = 1.

$z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(1)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(1)\pi}{4}))$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{2\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{2\pi}{4}))\\$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{2}))\\$

Root #2:

$\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{7\pi}{9}))+\mathfrak{i}(sin(\frac{7\pi}{9}))}$

Part 4: Solving for root #3

To solve for root #3, follow the same simplifying steps above but change k to k = 2.

$z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(2)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(2)\pi}{4}))$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{4\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{4\pi}{4}))\\$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\pi))+\mathfrak{i}(sin(\frac{5\pi}{18}+\pi))\\$

Root #3:

$\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{23\pi}{18}))+\mathfrak{i}(sin(\frac{23\pi}{18}))}$

Part 4: Solving for root #4

To solve for root #4, follow the same simplifying steps above but change k to k = 3.

$z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(3)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(3)\pi}{4}))$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{6\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{6\pi}{4}))\\$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{3\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{3\pi}{2}))\\$

Root #4:

$\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{16\pi}{9}))+\mathfrak{i}(sin(\frac{16\pi}{19}))}$

The fourth roots of 16(cos 200° + i(sin 200°) are listed above.

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3 years ago

How many lines of symmetry does the letter r have

Gnesinka [82]

The letter R has no lines of symmetry.

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2 years ago

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Delvig [45]

Answer:

Step-by-step explanation:

I counted each of the cubes and got 64.

Then I divided it by 4 since you need to divide it be 1/4.

once I divided 4 for 64 I got 16.

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