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3 years ago

How to determine the maximum and minimum lengths of a missing side on a triangle?

1 answer:

6 0

Well to find the missing side you're going to need the other two. Take the one side and multiply by itself. Take the other one and multiply by itself. Add those two products. Then root it. Basically find something that multiplys by itself to get that product

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The graph of f(x) is show below.

vlada-n [284]

ANSWER

(2,2)

EXPLANATION

Since f(x) and its inverse function are symmetric about the line y=x, if (a,b) lies on the graph of f, then (b,a) must lie on f inverse.

The point (2,2) lies on f and the same time on f inverse.

The solution is a point that satisfies both equations.

Hence the correct choice is:

(2,2)

5 0

3 years ago

Read 2 more answers

The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

$\bar x = 260.1615$

$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

8 0

2 years ago

Help help help help me pls !,!,!,

Len [333]

Answer:

Step-by-step explanation:

First, you find the area of RPQ using 1/2×base×height. That is, 1/2×6×8. And the answer is 24. Then you multiply 24 by 4 to get the area of all the sides excluding the base to get 96.

8 0

3 years ago

Help me, I don't understand this problem

Crank

Read some notes on reflection. If it asks to reflect it on the x-axis youre going to reflect it either on the bottom or top plane. If it asks to reflect on y-axis youre going to reflect it on either the left or right plane.

5 0

3 years ago

Simplify 45/150 show work please:)

Jlenok [28]

Answer:

$\frac{3}{10}$