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sp2606 [1]
3 years ago
6

What is the area of the triangle in the coordinate plane?

Mathematics
2 answers:
xeze [42]3 years ago
8 0

Answer:

The area of the triangle is 16\ units^{2}

Step-by-step explanation:

we know that

The area of a triangle is equal to

A=\frac{1}{2}bh

where

b is the base of triangle

h is the height of triangle

In this problem

Observing the graph

b=(12-8)=4\ units

h=(10-2)=8\ units

substitute the values

A=\frac{1}{2}(4)(8)=16\ units^{2}

katovenus [111]3 years ago
6 0
To solve for the area of a triangle you use the formula A=b*h/2. The base for this triangle is 4 and the height is 8. Take 4*8 which is 32 and divide that answer in half to get 16 which is your answer.
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9.12cm<br> 10.36cm<br> 11.34cm<br> 12.61cm
Vinvika [58]
Tan80 = QR / RS
QR = 5.67 (2)
QR = 11.34

answer

<span>11.34cm</span>
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3 years ago
Which fraction is equal to ⅛?<br> ⁴⁄₆ ⁴⁄₃₀ ²⁄₁₆
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Answer:

2/16

Step-by-step explanation:

1/8 * 2/2 = 2/16

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Step-by-step explanation:

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You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
FrozenT [24]

Question:

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of ​$114.00. Assume the population standard deviation is ​$15.30. Construct a​ 90% confidence interval for the population mean.

Answer:

At the 90% confidence level, confidence interval = 110.2484 < μ < 117.7516

At the 95% confidence level, confidence interval = 109.53 < μ < 118.48

The 95% confidence interval is wider

Step-by-step explanation:

Here, we have

Sample size, n = 45

Sample mean, \bar x = $114.00

Population standard deviation, σ = $15.30

The formula for Confidence Interval, CI is given by the following relation;

CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}

Where, z is found for the 90% confidence level as ±1.645

Plugging in the values, we have;

CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}

or CI: 110.2484 < μ < 117.7516

At 95% confidence level, we have our z value given as z = ±1.96

From which we have CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}

Hence CI: 109.53 < μ < 118.48

To find the wider interval, we subtract their minimum from the maximum as follows;

90% Confidence level: 117.7516 - 110.2484 = 7.5

95% Confidence level: 118.47503 - 109.5297 = 8.94

Therefore, the 95% confidence interval is wider.

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3 years ago
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