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2 years ago

11

tom was using wire of the following thicknesses .33mm , .275mm, .25mm, .3mm for some electrical work. order the wire from thicke

st to thinest

1 answer:

Lena [83]2 years ago

5 0

Thickest to thinnest....
0.33m , 0.3mm, 0.275mm, 0.25mm

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The summer after sophomore year, Haley nannied for a family and earned $2000. She

AnnZ [28]

You want to calculate the interest on $2000 at5.8% interest per month after six years?

Here is your formula: I =p*r*t

P is the principal amount which is $2000

R is the rate of interest which is 5.8% per month

T is the time involved whihc is six years

You’re interest is 8352.00

3 0

2 years ago

Which are the solutions of x2 = –13x – 4? 0, 13 0, –13

masha68 [24]

It's defiantly not C. I made a 90% but missed this question. I hope this helps you cross it down. My next guess would be D.

3 0

2 years ago

Read 2 more answers

Marisol is making a rectangular wooden frame. She wants the length of the frame to be no more than 12 inches. She has less than

svlad2 [7]

The perimeter of a rectangle is
2l + 2w
This has to be less than 30 because she has less than 30 inches of wood altogether.
The length must also be "no more than", which means less than or equal to, 12 inches.
Option A satisfies both of these conditions.

5 0

1 year ago

A toy factory makes 5,000 teddy bears per day. The supervisor randomly selects 10 teddy bears from all 5,000 teddy bears and use

tankabanditka [31]

Answer:

9

Step-by-step explanation:

<em>Degrees of freedom</em> is the number of values in the final calculation of a statistic that are free to vary. Degrees of freedom are related to sample size and calculated by n-1 where n is the <em>sample size</em>.

The Supervisor selects 10 teddy bears as sample from 5000 teddy bears produced daily. Therefore in this situation, there are 10-1=9 degrees of freedom

3 0

2 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me

Reil [10]

Answer:

a) $\bar X=10.65$

$Median =\frac{10.7+10.7}{2}=10.7$

$Mode= 10.7$

b) $Range = 11.8-8.3=3.5$

$s= 0.948$

c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

$s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0

2 years ago