Find the solution of the system of equations shown on the graph.

X^2 + y^2 less than or equal to 16 Please Graph with shaded areas

Ivanshal [37]

Answer:

See the diagram.

Step-by-step explanation:

$x^{2} +x^{2} \leq16$ represents a circular disc with centre $(0,0)$ and radius $4$

7 0

4 years ago

A²,b²,c² are consecutive perfect squares.How many natura numbers are lying between a² and c², if a>0

pogonyaev

Answer:

The quantity of natural numbers between $a^{2}$ and $c^{2}$ is $2\cdot (a + b) + 1$ .

Step-by-step explanation:

If $a^{2}$ , $b^{2}$ and $c^{2}$ are consecutive perfect squares, then both $a$ , $b$ and $c$ are natural numbers and we have the following quantities of natural numbers:

Between $b^{2}$ and $c^{2}$ :

$c^{2} = (b+1)^{2}$

$c^{2} = b^{2}+2\cdot b + 1$

$c^{2}-b^{2} = 2\cdot b + 1$

And the quantity of natural numbers between $b^{2}$ and $c^{2}$ is:

$c^{2}-b^{2}-1 = 2\cdot b$

Between $a^{2}$ and $b^{2}$ :

$b^{2} = (a + 1)^{2}$

$b^{2} = a^{2} +2\cdot a + 1$

$b^{2}-a^{2} = 2\cdot a + 1$

And the quantity of natural numbers between $a^{2}$ and $b^{2}$ is:

$b^{2}-a^{2}-1 = 2\cdot a$

And the quantity of natural numbers between $a^{2}$ and $c^{2}$ is:

$Diff = 2\cdot a + 2\cdot b + 1$

Please observe that the component +1 represents the natural number $b^{2}$

4 0

3 years ago

Construct a frequency distribution and a relative frequency distribution for the light bulb data with a class width of 20, start

k0ka [10]

Answer:

Step-by-step explanation:

Hello!

You have the information about light bulbs (i believe is their lifespan in hours) And need to organize the information in a frequency table.

The first table will be with a class width of 20, starting with 800. This means that you have to organize all possible observations of X(lifespan of light bulbs) in a class interval with an amplitude of 20hs and then organize the information noting their absolute frequencies.

Example

1) [800;820) only one observation classifies for this interval x= 819, so f1: 1

2)[820; 840) only one observation classifies for this interval x= 836, so f2: 1

3)[840;860) no observations are included in this interval, so f3=0

etc... (see attachment)

[ means that the interval is closed and starts with that number

) means that the interval is open, the number is not included in it.

fi: absolute frequency

hi= fi/n: relative frequency

To graph the histogram you have to create the classmark for each interval:

x'= (Upper bond + Lower bond)/2

As you can see in the table, there are several intervals with no observed frequency, this distribution is not uniform least to say symmetric.

To check the symmetry of the distribution is it best to obtain the values of the mode, median and mean.

To see if this frequency distribution has one or more modes you have to identify the max absolute frequency and see how many intervals have it.

In this case, the maximal absolute frequency is fi=6 and only one interval has it [1000;1020)

$Mo= LB + Ai (\frac{D_1}{D_1+D_2} )\\$

LB= Lower bond of the modal interval

D₁= fmax - fi of the previous interval

D₂= fmax - fi of the following interval

Ai= amplitude of the modal interval

$Mo= 1000 + 20*(\frac{(6-3)}{(6-3)+(6-4)} )=1012$

This distribution is unimodal (Mo= 1012)

The Median for this frequency:

Position of the median= n/2 = 40/2= 20

The median is the 20th fi, using this information, the interval that contains the median is [1000;1020)

$Me= LB + Ai*[\frac{PosMe - F_{i-1}}{f_i} ]$

LB= Lower bond of the interval of the median

Ai= amplitude of the interval

F(i-1)= acumulated absolute frequency until the previous interval

fi= absolute frequency of the interval

$Me= 1000+ 20*[\frac{20-16}{6} ]= 1013.33$

Mean for a frequency distribution:

$X[bar]= \frac{sum x'*fi}{n}$

∑x'*fi= summatory of each class mark by the frequency of it's interval.

∑x'*fi= (810*1)+(230*1)+(870*0)+(890*2)+(910*4)+(930*0)+(950*4)+(970*1)+(990*3)+(1010*6)+(1030*4)+(1050*0)+(1070*3)+(1090*2)+(1110*4)+(1130*0)+(1150*2)+(1170*1)+(1190*1)+(1210*0)+(1230*1)= 40700

$X[bar]= \frac{40700}{40} = 1017.5$

Mo= 1012 < Me= 1013.33 < X[bar]= 1017.5

Looking only at the measurements of central tendency you could wrongly conclude that the distribution is symmetrical or slightly skewed to the right since the three values are included in the same interval but not the same number.

*-*-*

Now you have to do the same but changing the class with (interval amplitude) to 100, starting at 800

Example

1) [800;900) There are 4 observations that are included in this interval: 819, 836, 888, 897 , so f1=4

2)[900;1000) There are 12 observations that are included in this interval: 903, 907, 912, 918, 942, 943, 952, 959, 962, 986, 992, 994 , so f2= 12

etc...

As you can see this distribution is more uniform, increasing the amplitude of the intervals not only decreased the number of class intervals but now we observe that there are observed frequencies for all of them.

Mode:

The largest absolute frequency is f(3)=15, so the mode interval is [1000;1100)

Using the same formula as before:

$Mo= 1000 + 100*(\frac{(15-12)}{(15-12)+(15-8)} )=1030$

This distribution is unimodal.

Median:

Position of the median n/2= 40/2= 20

As before is the 20th observed frequency, this frequency is included in the interval [1000;1100)

$Me= 1000+ 100*[\frac{20-16}{15} ]= 1026.67$

Mean:

∑x'*fi= (850*4)+(950*12)+(1050*15)+(1150*8)+(1250*1)= 41000

$X[bar]= \frac{41000}{40} = 1025$

X[bar]= 1025 < Me= 1026.67 < Mo= 1030

The three values are included in the same interval, but seeing how the mean is less than the median and the mode, I would say this distribution is symmetrical or slightly skewed to the left.

I hope it helps!

8 0

3 years ago

A business owner collected data on the people working for her company. She determined the average number of minutes each employe

Naddika [18.5K]

Answer:

The comparison using median and IQR is best because one of the graphs is not symmetrical.

Step-by-step explanation:

The following information is missing

A box plot titled Number of Minutes Women Spend on Breaks. The number line goes from 30 to 70. The whiskers range from 30 to 54, and the box ranges from 34 to 50. A line divides the box at 48.

A box plot titled Number of Minutes Men Spend on Breaks. The number line goes from 30 to 70. The whiskers range from 30 to 68, and the box ranges from 36 to 60. A line divides the box at 48.

The business owner uses the median and IQR to determine the center and variability of the data sets. Which best describes the comparison?

The comparison would be more accurate using the mean and MAD because one of the graphs is symmetric.

The comparison would be more accurate using the mean and MAD because the median of both data sets is the same.

The comparison using median and IQR is best because one of the graphs is not symmetrical.

The comparison using median and IQR is best because the median is greater than the IQR for both data sets.

Mean and MAD are useful for comparison when both data sets are symmetrical.

In the women box plot the Q1 is at 34, the median is at 48, and the Q3 is at 50, so it is not symmetrical (the difference between the median and the Q1, and the Q3 and the median is not the same)

8 0

4 years ago

Math interest, I, P, R,T I will give more points to you once questions are completed.

anyanavicka [17]

The interest based on the information given is $2564.

<h3>How to calculate the interest?</h3>

From the information given, the following can be deduced:

Principal = $36900

Time = 178 days

Rate = 14.25%

The simple interest will be:

= (PRT)/100

= (36900 × 14.25 × 178/365)/100

= $2564

Learn more about interest on:

brainly.com/question/26011426

#SPJ1

4 0

2 years ago

Find the solution of the system of equations shown on the graph.​

Find the solution of the system of equations shown on the graph.