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nikdorinn [45]
3 years ago
13

A nurse collected data about the average birth weight of babies in the hospital that month. Her data is shown using the dot plot

. Create a box plot to represent the data. dot plot titled Monthly Birth Weight and number line from 8 to 9 in increments of 1 tenth labeled Birth Weight (in pounds.) with zero dots over 8, zero dots over 8 and 1 tenth, 2 dots over 8 and 2 tenths, 2 dots over 8 and 3 tenths, 2 dots over 8 and 4 tenths, 3 dots over 8 and 5 tenths, zero dots over 8 and 6 tenths, 1 dot over 8 and 7 tenths, zero dots over 8 and 8 tenths, 1 dot over 8 and 9 tenths, and zero dots over 9 box plot with minimum value 8 and 3 tenths, lower quartile 8 and 4 tenths, median 8 and 5 tenths, upper quartile 8 and 6 tenths, and maximum value 8 and 9 tenths box plot with minimum value 8 and 2 tenths, lower quartile 8 and 3 tenths, median 8 and 4 tenths, upper quartile 8 and 5 tenths, and maximum value 8 and 9 tenths box plot with minimum value 8 and 3 tenths, lower quartile 8 and 5 tenths, median 8 and 6 tenths, upper quartile 8 and 7 tenths, and maximum value 8 and 9 tenths box plot with minimum value 8 and 2 tenths, lower quartile 8 and 3 tenths, median 8 and 4 tenths, upper quartile 8 and 6 tenths, and maximum value 8 and 9 tenths
Mathematics
1 answer:
mixer [17]3 years ago
5 0

Answer:

The values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

Step-by-step explanation:

The data provided is as follows:

X     Frequency

8        0

8.1        0

8.2        2

8.3        2

8.4        2

8.5        3

8.6        0

8.7         1

8.8        0

8.9         1

 9           0

So, the actual data is:

S = {8.2 , 8.2 , 8.3 , 8.3 , 8.4 , 8.4 , 8.5 , 8.5 , 8.5 , 8.7 , 8.9 }

A boxplot, also known as a box and whisker plot is a method to demonstrate the distribution of a data-set based on the following 5 number summary,

  1. Minimum (shown at the bottom of the chart)
  2. First Quartile (shown by the bottom line of the box)
  3. Median (or the second quartile) (shown as a line in the center of the box)
  4. Third Quartile (shown by the top line of the box)
  5. Maximum (shown at the top of the chart).

The data set is arranged in ascending order.

The minimum value is, Min. = 8.2

The lower quartile is,

[\frac{n+1}{4}]^{th}\ obs.=[\frac{11+1}{4}]^{th}\ obs.=3^{rd }\ obs. =8.3,

Q₁ = 8.3.

The median value is,

[\frac{n+1}{2}]^{th}\ obs.=[\frac{11+1}{2}]^{th}\ obs.=6^{th}\ obs.=8.4

Median = 8.4

The upper quartile is,

[\frac{3(n+1)}{4}]^{th}\ obs.=[\frac{3(11+1)}{4}]^{th}\ obs.=9^{th }\ obs. =8.5,

Q₃ = 8.5.

The maximum value is, Max. = 8.9.

So, the values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

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at Michaels auto shop, it takes him 15 minutes to do an oil change and 24 minutes to do a tire change. let x be the number of oi
Roman55 [17]

Answer:

Step-by-step explanation:

At Susan's auto shop, it takes her 12 minutes to do an oil change and 18 minutes to do a tire change. Let x be the number of oil changes she does. Let y be the number of tire changes she does.

Using the values and variables given, write an inequality describing how many oil changes and tire changes Susan can do in less than 3 hours (180 minutes).

Answer provided by our tutors

The total time that Susan needs to change x oil changes and y tire changes is less than 180 min.

The time needed for x oil changes is 12 * x.

The time needed for y tire changes is 18 * y.

The total time is the sum of the above times and needs to be less than 180 that is

12 * x + 18 * y < 180 divide both sides of equation by 6

12/6 * x + 18/6*y < 180/6

2*x + 3*y < 30

2*x < 30 - 3*y divide both sides by 2 to get the inequality for x

x < 30/2 - 3/2*y = 15 - 1.5 y < 15 that is x < = 15

2*x + 3*y < 30

3*y < 30 - 2*x divide both sides by 3 to get the inequality for y

y < 30/3 - 2/3 *x = 10 - 2/3*x < 10 that is y < = 10

Also we can write x + y < x+ 3/2 * y < 15.

Susan can do not more then 5 oil changes and not more then 10 tire changes or all together she can do not more then 15 total of oil and tire changes.

Just do this problem but with Michael

8 0
2 years ago
8x+2y=-2<br> y=-5x+1<br> Solve for x and y
sp2606 [1]
X = 2
y = -9
16 - negative 18 = -2

5 0
3 years ago
Read 2 more answers
(25 points) Can someone please solve this I just need to see how its solved to understand
aleksley [76]

x = total amount of students in 8th Grade.

we know only one-thrid of the class went, so (1/3)x or x/3 went.

we also know 5 coaches went too, and that the total amount of that is 41.

\bf \stackrel{\textit{one third of all students}}{\cfrac{1}{3}x}+\stackrel{\textit{coaches}}{5}=\stackrel{\textit{total}}{41}\implies \cfrac{x}{3}+5=41\implies \cfrac{x}{3}=41-5 \\\\\\ \cfrac{x}{3}=36\implies x=3(36)\implies x=108

now, to verify, well, what do you get for (108/3) + 5?

4 0
3 years ago
Triangle EFG is dilated by a scale factor of 3 centered at (0, 1) to create triangle E'F'G'. Which statement is true about the d
jolli1 [7]

Answer:

Segment EH and segment E prime H prime both pass through the center of dilation (A)

The complete question related to this found on brainly (ID:16812154) is stated below:

Triangle EFG is dilated by a scale factor of 3 centered at (0, 1) to create triangle E'F'G'. Which statement is true about the dilation?

E(0,5) F(1,1) G(-2,1) H(0,1)

a) segment EH and segment E prime H prime both pass through the center of dilation.

b)The slope of segment EF is the same as the slope of segment E prime H prime.

c) segment E prime G prime will overlap segment EG. segment

d) EH ≅ segment E prime H prime.

Step-by-step explanation:

∆EFG = Original image

∆EFG is dilated to give ∆E'F'G'

∆E'F'G' = New image

Scale factor = 3

Center of dilation = (0,1) = H(0,1)

Coordinates ∆EFG : E(0,5) F(1,1) G(-2,1)

To determine the statement that is true about the dilation from the options,

First we would make a diagram on the coordinates of ∆EFG and center of dilation (H).

Find attached the diagram.

Length GF = 3unit

Length G'F' = 3×scale factor = 9unit

Length EH = 4unit

Length E'H' = 4×scale factor = 12unit

Then we would move 3units to the left on same line from G to get the coordinate of G'(mark the point).

Also move 3units to the right on same line from F to get the coordinate of F'(mark the point).

Both of these give length of G'F' = 9unit

Now move 8units to the top from E to get the coordinate of E'(mark the point).

From this you get length E'H' = 12unit

Draw lines connecting the three points to get ∆E'F'G'

See diagram for better understanding

From the diagram, EH and E'H' both pass through (0,1). The other options are wrong.

Therefore, Segment EH and segment E prime H prime both pass through the center of dilation (A)

5 0
3 years ago
Find the area of the triangle
BARSIC [14]

Answer:

The Area of Δ ABC = 219.13

Step-by-step explanation:

<em>The hard part about this problem is finding the area without the height</em>

The formula to do this is Area = \sqrt{S(S-A)(S-B)(S-C)}

A, B, C represent the sides

S represents \frac{1}{2} (A + B + C)

In this equation, we will make the base be A, and the other two sides will be B and C

<u>Sides B and C are the same length</u> because they meet at a 90° angle

Lets plug the numbers into the variables

A = 28

B= 21

C= 21

<u>Remember:</u> S represents  \frac{1}{2}  (A + B + C)

S =  \frac{1}{2} (28 + 21 + 21)

S =  \frac{1}{2} (70)

S = 35

<em>Lets plug the numbers into the Area Formula now!</em>

Area = \sqrt{35(35 - 28)(35 - 21)(35 - 21)}

According to the order of operations, we need to do the calculations in parentheses <u>first</u>

35 - 28 = 7

35 - 21 = 14

35 - 21 = 14

14 x 14 x 7 = 1372

1372 x 35 = 48020

\sqrt{48020} = 219.13

The Area of Δ ABC = 219.13

3 0
3 years ago
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