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Elodia [21]
4 years ago
11

What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?

Mathematics
2 answers:
Alexxx [7]4 years ago
7 0

Answer:the answer is 21.6.


Step-by-step explanation:


Leno4ka [110]4 years ago
3 0
To calculate the perimeter, consider each side separately.

Right side: This is easiest since it's a vertical line - it's 7 units long.

Top side: If you look carefully, this side is really the hypotenuse of a triangle that is 1 unit tall and 6 units long. Using the Pythagorean Theorem, you can calculate the length of the hypotenuse to be 1^2 + 6^2 = c^2 --> c = 6.1

Left side: just like the top side, this side is the hypotenuse of a triangle that is 6 units tall and 6 units long. Pythagoras again: 6^2 + 6^2 = c^2 --> c = 8.5

Add these three numbers to get the perimeter: 7 + 6.1 + 8.5 = 21.6
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Using division we get
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(19 POINTS QUESTION, HELP PLEASE) A bag contains 5 blue balls, 4 red balls, and 3 orange balls. If a ball is picked from the bag
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B.  0.4

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5 + 4 + 3 = 12 balls

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Tell whether the angles are complementary or supplementary. Then find the value of x.
Margaret [11]

Answer: Complementary angles. x = 60

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5 0
3 years ago
Some transportation experts claim that it is the variability of speeds, rather than the level of speeds, that is a critical fact
scZoUnD [109]

Answer:

Explained below.

Step-by-step explanation:

The claim made by an expert is that driving conditions are dangerous if the variance of speeds exceeds 75 (mph)².

(1)

The hypothesis for both the test can be defined as:

<em>H</em>₀: The variance of speeds does not exceeds 75 (mph)², i.e. <em>σ</em>² ≤ 75.

<em>Hₐ</em>: The variance of speeds exceeds 75 (mph)², i.e. <em>σ</em>² > 75.

(2)

A Chi-square test will be used to perform the test.

The significance level of the test is, <em>α</em> = 0.05.

The degrees of freedom of the test is,

df = n - 1 = 55 - 1 = 54

Compute the critical value as follows:

\chi^{2}_{\alpha, (n-1)}=\chi^{2}_{0.05, 54}=72.153

Decision rule:

If the test statistic value is more than the critical value then the null hypothesis will be rejected and vice-versa.

(3)

Compute the test statistic as follows:

\chi^{2}=\frac{(n-1)\times s^{2}}{\sigma^{2}}

    =\frac{(55-1)\times 94.7}{75}\\\\=68.184

The test statistic value is, 68.184.

Decision:

cal.\chi^{2}=68.184

The null hypothesis will not be rejected at 5% level of significance.

Conclusion:

The variance of speeds does not exceeds 75 (mph)². Thus, concluding that driving conditions are not dangerous on this highway.

7 0
3 years ago
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