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11Alexandr11 [23.1K]
3 years ago
13

Solve the problem in the image attached

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
8 0
I can still help you but there no image?
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Find the slope of the line that passes through the pair of points.<br><br> (–5.5, 6.1), (–2.5, 3.1)
hjlf

Slope of the line is -1.

Step-by-step explanation:

Given points are,

(-5.5,6.1),(-2.5,3.1)

Here,

x_{1} = -5.5 ,\ \ \ x_{2}=-2.5\\y_{1} = 6.1,\ \ \ y_{2}=3.1

Slope = m = \frac{y_{2}-y_{1}}{x_{2}-x{1}}

Putting values;

m=\frac{3.1-6.1}{-2.5-(-5.5)}\\m=\frac{-3}{-2.5+5.5}\\m=\frac{-3}{3}\\m=-1

Slope of the line is -1.

Keywords: Slope, line

Learn more about slopes at:

  • brainly.com/question/2977815
  • brainly.com/question/3071107

#LearnwithBrainly

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A chef mixed oil and vinegar to make a salad dressing she made enough dressing to fill 6 plastic bottles each bottle has 4/8 cup
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48 cups because there are 8 cups in a bottle and 8 times 6 is 48
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Finding missing angles
Bingel [31]

Answer:

D)

Step-by-step explanation:

Since we see the whole angle is 90° and it's split into two smaller angles, 7x° and 11x°, then we can conclude that 7x°+11x°=90°

*And x=5

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If it takes 6 hours to mow 3 lawns how many. lawns could he mow in 36 hours​
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Read 2 more answers
An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th
Natasha2012 [34]

Answer:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

Step-by-step explanation:

Notation

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

For this case the 9% confidence interval is given by:

8.8104 \leq \mu \leq 11.1248

We can calculate the mean with the following:

\bar X = \frac{8.8104 +11.1248}{2}= 9.9676

And we can find the margin of error with:

ME= \frac{11.1248- 8.8104}{2}= 1.1572

The margin of error for this case is given by:

ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE

And we can solve for the standard error:

SE = \frac{ME}{t_{\alpha/2}}

The critical value for 95% confidence using the normal standard distribution is approximately 1.96 and replacing we got:

SE = \frac{1.1572}{1.96}= 0.5904

Now for the 98% confidence interval the significance is \alpha=1-0.98= 0.02 and \alpha/2 = 0.01 the critical value would be 2.326 and then the confidence interval would be:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

8 0
3 years ago
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