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2 years ago

<1 and <2 are _____ angles? - verticals - complementary -adjacent -obtuse

Mathematics

2 answers:

patriot [66]2 years ago

5 0

Answer:

<1 and <2 are adjacent

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They arent equal so its not vertical

They arent complementary because they dont add up to 90

They are adjacent because they share a common side <--

They arent obtuse because <1 is acute

GarryVolchara [31]2 years ago

3 0

Answer:

Complementary

Step-by-step explanation:

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2 years ago

(10.4)2 =<br> to the second power is the 2

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Answer:

10.4^2 = 108.16

Step-by-step explanation:

10.4^2 = 10.4 * 10.4

10.4 * 10.4 = (10 * 10.4) + (0.4 * 10.4)

(10 * 10.4) + (0.4 * 10.4) = (104) + (4.16)

(104) + (4.16) = 108.16

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2 years ago

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The Big Telescope Company sells circular mirrors. Their largest mirrors have radii of 5 meters and their smallest mirrors have r

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Answer: 1:5

Step-by-step explanation:

Given: The cost of every mirror is proportional to the cube of the mirror's radius.

i.e. $\dfrac{\text{Cost of smallest mirror}}{\text{Cost of largest mirror}}=\dfrac{(\text{radii of smallest mirror}^3)}{\text{(radii of largest mirror)}^3}$

Their largest mirrors have radii of 5 meters and their smallest mirrors have radii of 1 meter.

Then,

$\dfrac{\text{Cost of smallest mirror}}{\text{Cost of largest mirror}}=\dfrac{1^3}{(5)^3}=\dfrac{1}{125}$

The ratio of the total cost of 25 of the company's smallest mirrors to the cost of one of the company's largest mirrors will be:

$\dfrac{\text{Cost of 25 smallest mirror}}{\text{Cost of largest mirror}}=\dfrac{25\times 1}{125}=\dfrac{1}{5}$

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2 years ago

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The

Tems11 [23]

Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that $n = 387, \pi = \frac{298}{387} = 0.77$

84% confidence level

So $\alpha = 0.16$ , z is the value of Z that has a p-value of $1 - \frac{0.16}{2} = 0.92$ , so $Z = 1.405$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8$

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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