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borishaifa
21 day ago
7

The Whitcomb Company manufactures a metal ring for industrial engines that usually weighs about 50 ounces. A random sample of 50

of these metal rings produced the following weights (in ounces). 51 53 56 50 44 47 53 53 42 57 46 55 41 44 52 56 50 57 44 46 41 52 69 53 57 51 54 63 42 47 47 52 53 46 36 58 51 38 49 50 62 39 44 55 43 52 43 42 57 49
(a) Construct the frequency table and histogram for these data using eight classes. (You can choose your own classes)
(b) What can you observe about the data from the histogram? tssabout 500 nces. Doyoub elieve their claim? Explain your reason.

Mathematics
2 answers:
Svet_ta [9.4K]21 day ago
8 0

It’s important to begin by creating a frequency table to simplify plotting, subsequently organizing your data into classes for ease of visualization, such as 35-40, 41-45, etc.

B) Observations reveal a significant demand for metal rings weighing between 50-55 ounces and from 55-60 ounces.

babunello [8.4K]21 day ago
3 0

Answers:

a) Please check the images below.

b) No.

Step-by-step explanation:

Hi,

A histogram resembles a bar chart but bins numbers into intervals.

The weights data signifies:

Range = Maximum Value - Minimum Value

= 69 - 36

= 33

Required classes are 8.

Class\ width = \frac{Range}{Number\ of\ classes}\\= \frac{33}{8}\\= 4.125

As we construct our data with this width, it will leave a few values out; hence, we slightly adjust the max and min values to ensure all values fit into the range.

Approximate max = 71

Approximate min = 35

New range = 71 - 35

= 36

New\ Class\ width = \frac{36}{8}\\= 4.5

Based on this, we formulate our frequency table:

Starting from 35, we add 4.5 to reach 39.5.

The next value is obtained by adding another 4.5 to 39.5, making it 44.0.

This continues until all classes are established:

35.0 - 39.5

39.5 - 44.0

44.0 - 48.5

48.5 - 53.0

53.0 - 57.5

57.5 - 62.0

62.0 - 66.5

66.5 - 71.0

With each class width, count how many values fall within those intervals to establish the frequency count.

The final table appears as displayed in the accompanying image below.

To build a histogram, plot the class intervals (x-axis) against the frequency (y-axis). The histogram will resemble the image below.

b)

A close inspection of the table indicates that frequencies are skewed more towards the left. Thus, we characterize this distribution as left-skewed.

Additionally, it can be inferred that the mean and other central values are pushed more towards the left side of the graph.

Hence, the claim that their metal rings weigh around 50 ounces seems unfounded as central values are predominantly left-skewed.

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A study1 conducted in July 2015 examines smartphone ownership by US adults. A random sample of 2001 people were surveyed, and th
Inessa [8997]

Answer:

a) Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}  

b) z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679  

c) z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58    

d) In this scenario, we notice that \hat p_1 > \hat p_2 thus the conclusion for this case would indicate

Step-by-step explanation:

Information provided

X_{1}=688 denote the number of men possessing smartphones  

X_{2}=671 signify the number of women possessing smartphones

n_{1}=989 group of men sampled

n_{2}=1012 group of women sampled

p_{1}=\frac{688}{989}=0.696 symbolize the proportion of men with smartphones

p_{2}=\frac{671}{1012}=0.663 symbolize the proportion of women with smartphones

\hat p denote the pooled estimate of p

z would denote the test statistic

p_v signify the value

Part a

The objective is to evaluate if there is a disparity in smartphone ownership between men and women; the hypothesis statements would be:  

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

Part b

The statistic relevant to this case is expressed as:

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679  

Part c

By substituting the provided information, we find:

z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58    

Part d

In this instance, it is evident that \hat p_1 > \hat p_2 thus the conclusion for this case would seem

4 0
5 days ago
"A sample of 20 randomly chosen water melons was taken from a large population, and their weights were measured. The mean weight
AnnZ [9099]

Answer: (97.98, 112.020)

Step-by-step explanation: We will create a 95% confidence interval for the average weight of melons.

Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

Sample mean = x = 105 lb

Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

tα/2 represents the critical value for the test (which will be determined using the t distribution table).

To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

<pfor the="" lower="" limit="" we="" find:="">

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

</pfor>
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23 days ago
Suppose R is plotted in the above coordinate plane and is collinear with A and B. If the ratio AR: AB is 1/3, then what are the
babunello [8412]
= -4/2 Step-by-step explanation: xp=x1 +k (x2-x1) = -3+1/3 (2- -3) = -3 + 1/3  (5/1) = -3 + 5/3 = -9/3 +5/3 = -4/2 Sorry, it doesn't look the best:)
3 0
14 days ago
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A series contains 18 numbers and has a sum of 4,185. The last number of the series is 275. What is the first number?
Inessa [8997]
The formula for the sum of an arithmetic series is expressed as:
Sn=n/2(a1+an)
where:
n=total terms
a1=the initial term
an=the final term
given
n=18, an=275, Sn=4185
substituting these values into the equation results in:
4185=18/2(a1+275)
after simplification, we obtain:
4185=9(a1+275)
dividing by 9 yields:
465=a1+275
therefore
a1=465-275
a1=190

Conclusion: the first term is 190
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11 days ago
The Sears Tower, at 1,451 feet, is one of the tallest structures in the United States. A penny is thrown from the top of the tow
Leona [9267]

Answer:

The formula representing the penny's height as a function of time is:

h(t)=1451-16t^2

After 7 seconds, the height of the penny will reach 667 feet.

Step-by-step explanation:

The penny experiences free fall.

With an initial velocity of zero and an initial height of h(0)=1,451.

Gravity acts as the acceleration, measured as g=32 ft/s^2.

The model can be initiated by analyzing speed:

dv/dt=-g\\\\v(t)=v_0-gt=-gt

Then, the height is expressed as:

dh/dt=v(t)=-gt\\\\h(t)=h_0-\dfrac{gt^2}{2}=1451-\dfrac{32}{2}t^2\\\\\\h(t)=1451-16t^2

The height of the penny at approximately 7 seconds can be calculated as:

h(7)=1451-16(7^2)=1451-16*49=1451-784=667

After 7 seconds, the penny will stand at a height of 667 feet.

6 0
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