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lapo4ka [179]
3 years ago
9

Evaluate x-(-y) when x= -2.31 and y = 5.9

Mathematics
2 answers:
Sonbull [250]3 years ago
7 0
Answer is:
3.59
How to solve:
1. Replace variables
2. Do equation
kondor19780726 [428]3 years ago
4 0
-2.31-(-5.9)= -2.31+5.9= 3.59
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HELPPP! Last attempt geometry!!
Helen [10]

Answer:

  x1 = 2, y1 = 0

  x2 = 2, y2 = 6

Step-by-step explanation:

Each pair of coordinates is (x, y). The first pair is (x1, y1). The second pair is (x2, y2). This means you have ...

  x1 = 2, y1 = 0

  x2 = 2, y2 = 6

_____

<em>Additional comment</em>

The slope of the line through these points is ...

  m = (y2 -y1)/(x2 -x1) = (6 -0)/(2 -2) = 6/0 = undefined

The line through these points is a vertical line with equation x = 2.

6 0
3 years ago
Solve the proportion: 5/x+7=3/x+3
Allushta [10]

Answer:

x=3

Step-by-step explanation:

cross multiply

5(x+3)=3(x+7)

5x+15=3x+21

5x+15-3x=3x+21-3x

2x+15=21

2x+15-15=21-15

2x=6

x=3

7 0
3 years ago
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
3 years ago
Please help me do this problem! ‼️
blagie [28]
0 - 18 = -18

-18 + 9 = -9 ft
7 0
3 years ago
TONTE UI E doove
vagabundo [1.1K]

pretty sure its growth

5 0
2 years ago
Read 2 more answers
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