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adoni [48]
3 years ago
8

Use the data set to answer the question.

Mathematics
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

The Mean Absolute Deviation for the data set is 46.5

Step-by-step explanation:

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yanalaym [24]

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7 0
3 years ago
If you answer these two questions, that would be awesome, on my other questions i have asked the same one twice so if you want m
Masja [62]

Answer:

\sqrt[3]{x^2}

x ^( 1/8)

Step-by-step explanation:

x ^ (5/6)  ÷ x ^ (1/6)

We know that a^ b ÷ a ^c = a^ ( b-c)

x^ ( 5/6 - 1/6)

x^ (4/6)

x ^ 2/3

\sqrt[3]{x^2}

\sqrt{x}\sqrt[\\4]{x}

x ^ 1/2 * x ^ 1/4

We know that a^ b * a ^ c = a^ (b*c)

x ^ (1/2 * 1/4)

x ^( 1/8)

3 0
3 years ago
Read 2 more answers
A jar of crunchy peanut butter contains 1.35 kg of peanut butter. if you use 8.0 % of the peanut butter for a sandwich, how many
Travka [436]
1. You have that:
  
 -The jar of crunchy peanut butter contains 1.35 kilograms of peanut butter. 
<span> 
 -You use 8.0 % of the peanut butter.
</span> 

 2. You must convert 1.35 kilograms to ounces, as below:
 

 1 kilogram=35.2739 ounces
 
 =(1.35 kilogram)(35.2739 ounces/1 kilogram)
 =1.35x35.2739 ounces
 =47.61 ounces
 

  3. Then, you have:
 
 47.61 ounces-----100%
                     x-----8%
 
 8%/100=0.08
 100%/100=1
 
 x=(0.08x47.61 ounces)/1
 x=3.80 ounces
 
 H<span>ow many ounces of peanut butter did you take out of the container?</span> 
 
 The answer is: 3.80 ounces 
6 0
3 years ago
Josh weighed cantaloupes he picked from his garden. This line plot shows the cantaloupe weights. What is the total weight of all
pashok25 [27]

Answer:

ip got a question how many cantaloupes are there

8 0
3 years ago
10.4.1 .WP The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at ran
Sunny_sXe [5.5K]

Answer:

[-0.65;2.54]km

Step-by-step explanation:

Hello!

So, you need to find a 99%CI for the difference in mean life of two brands of radial tires. Since he assigned one tire of each brand at random to the two rear wheels of each car, in other words, every car tested had one rear tire of each brand at the same time, this test can be considered to be of paired samples.

           Brand 1    Brand 2    X₁-X₂

car 1:    36,925 ;   34,318   ; 2.607

car 2:   45,300 ;   42,280  ; 3.020

car 3:   36,240 ;    35,500 ; 0.740

car 4:   32,100  ;    31,950  ; 0.150

car 5:   37,210  ;    38,015  ; -.0805

car 6:   48,360 ;    47,800 ; 1.160

car 7:   38,200 ;    37,810  ; 0.390

car 8:   33,500 ;    33,215  ; 0.285

n= 8

With this in mind, we define the study variable as Xd= X₁-X₂

Where X₁ corresponds to the lifespan, in km, of a tire from Brand 1

and X₂ corresponds to the lifespan, in km, of a tire from Brand 2

so Xd will be the difference between the lifespan of the tires from Brand 1 and Brand 2.

This variable Xd~N(μd;δd²) (p-value for normality test is 0.4640)

To calculate the CI the best statistic is the Student's t with the following formula:

t= (xd[bar] - μd)/(Sd/√n) ~t₍ₙ₋₁₎

sample mean: xd[bar]= 0.94

standard deviation: Sd= 1.29

t_{8;0.995} = 3.355

xd[bar] ± t_{8;0.995}*(Sd/√n)

⇒ 0.94 ± 3.355*(1.29/√8)

[-0.65;2.54]km

The Confidence Interval can be compared to a pair of bilateral hypothesis. If we were to determine the following hypothesis

H₀:μd=0

H₁:μd≠0

Using the level of significance of 0.01 (complementary to the confidence level)

As the calculated confidence interval contains zero, we do not reject the null hypothesis, that is, there is no significant difference between the two tire brands.

I hope you have a SUPER day!

5 0
3 years ago
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