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s2008m [1.1K]
4 years ago
12

Y = –x + 4 x + 2y = –8 How many solutions does this linear system have?

Mathematics
2 answers:
masya89 [10]4 years ago
4 0
It only has one solution. x= 16 and y = -12
Daniel [21]4 years ago
4 0
I just took the test, it's no solution. C
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What value of x makes the equation below true? 6x − 7 = 26 <br> A.5<br> B.5.5<br> C.6<br> D.6.5
makkiz [27]

Answer:

B

Step-by-step explanation:

6x-7=26

6x=33

x=5.5

3 0
3 years ago
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A plane flies from Houston to nyc in 4hours the distance it travelled was 1627miles what was the average speed of the plane
Tamiku [17]
406.75 since the equation for speed is distance over time so you would do 1627 divided by 4
8 0
4 years ago
Construct the indicated confidence interval for the population mean using the​ t-distribution. Assume the population is normally
DanielleElmas [232]

Using the t-distribution, it is found that the confidence interval is (14.2, 14.8).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm t\frac{s}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • t is the critical value.
  • n is the sample size.
  • s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a <em>two-tailed 95% confidence interval</em>, with 18 - 1 = <em>17 df</em>, is t = 2.1098.

The other parameters are given by:

\overline{x} = 14.5, s = 0.68, n = 18.

Hence, the bounds of the interval are given by:

\overline{x} - t\frac{s}{\sqrt{n}} = 14.5 - 2.1098\frac{0.68}{\sqrt{18}} = 14.2

\overline{x} + t\frac{s}{\sqrt{n}} = 14.5 + 2.1098\frac{0.68}{\sqrt{18}} = 14.8

The confidence interval is (14.2, 14.8).

More can be learned about the t-distribution at brainly.com/question/16162795

4 0
3 years ago
The following data represent the pulse rates? (beats per? minute) of nine students enrolled in a statistics course. Treat the ni
charle [14.2K]

Answer:

(a)77.4bpm

(b)Mean of Sample 1 = 70.3 beats per minute.  

Mean pulse of sample 2 = 70 beats per minute.

(c)

  • The mean pulse rate of sample 1 underestimates the population mean.
  • The mean pulse rate of sample 2 underestimates the population mean.

Step-by-step explanation:

(a)Population mean pulse.

The pulse of the nine students which represent the population are:

  • Perpectual Bempah      64
  • Megan Brooks               77
  • Jeff Honeycutt               89
  • Clarice Jefferson           69
  • Crystal Kurtenbach       89
  • Janette Lantka              65
  • Kevin McCarthy             88
  • Tammy Ohm                  69
  • Kathy Wojdya                 87

\text{Population Mean} =\dfrac{64+77+89+69+89+65+88+69+87}{9} \\=\dfrac{697}{9} \\\\=77.44

The population mean pulse is approximately 77.4 beats per minute.

(b)Sample 1: {Janette,Clarice,Megan}

  • Janette: 65bpm
  • Clarice: 69bpm
  • Megan: 77bpm

Mean of Sample 1

\text{Sample 1 Mean} =\dfrac{65+69+77}{3} \\=\dfrac{211}{3} \\\\=70.3

Sample 2: {Janette,Clarice,Megan}

  • Perpetual: 64bpm
  • Clarice: 69bpm
  • Megan: 77bpm

Mean of Sample 2

\text{Sample 2 Mean} =\dfrac{64+69+77}{3} \\=\dfrac{210}{3} \\\\=70

The mean pulse of sample 1 is approximately 70.3 beats per minute.  

The mean pulse of sample 2 is approximately 70 beats per minute.

(c)

  • The mean pulse rate of sample 1 underestimates the population mean.
  • The mean pulse rate of sample 2 underestimates the population mean.
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Kendra plans to go to the gym on 15 different days this month.
dangina [55]

Answer:4 answer

Step-by-step explanation:

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