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Yanka [14]
3 years ago
6

I need help. Include work plz

Mathematics
1 answer:
Dvinal [7]3 years ago
6 0

Answer:

Set Y has the smaller mean. Mean means average.

Step-by-step explanation:

Set X: 20.45+28+22.25+16.5+13+18.05=118.25 divide that by 6. 118.25/6= 19.71

Set Y: 21+6+5+24+9+40= 95 divide that by 6. 95/6= 15.83

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Artist 52 [7]

Answer:

I think d is the answer......

4 0
2 years ago
Read 2 more answers
Gavin needs $80 to buy a fish tank. He has saved $8 and plans to work as a babysitter to earn $9 per hour. Which inequality show
yuradex [85]
Gavin needs $80 to buy a fish tank. He has saved $8 and plans to work as a babysitter to earn $9 per hour. Which inequality shows the minimum number of hours, n, that Gavin should work as a babysitter to earn enough to buy the fish tank? 8 + 9n ≥ 80, so n ≥ 8 8 + 9n ≤ 80, so n ≤ 8 9n ≥ 80 + 8, so n ≥ 9.8 9n ≤ 80 + 8, so n ≤ 9.8
3 0
3 years ago
A rectangle has an area of 40 square units. The length is 6<br> units greater than the width.
stiks02 [169]

Answer:

Answer:

The width is 4 units, and the length is 10 units.

Step-by-step.

Step-by-step explanation:

area of rectangle = length * width

Let L = length; let W = width.

"The length is 6 units greater than the width.": L = W + 6

area = LW = 40

Since L = W + 6, we substitute L with W + 6.

(W + 6)W = 40

W^2 + 6W = 40

W^2 + 6W - 40 = 0

(W - 4)(W + 10) = 0

W - 4 = 0 or W + 10 = 0

W = 4 or W = -10

A width cannot be a negative number, so we discard the solution W = -10.

W = 4

L = W + 6 = 4 + 6 = 10

The width is 4 units, and the length is 10 units.

3 0
3 years ago
Read 2 more answers
5.8, 6<br> What is the area of triangle ABC?<br> I can’t figure out the answer
guapka [62]
You would need the 3rd side length of the triangle, in order to find the area.
4 0
3 years ago
Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul
natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

\bf B=\displaystyle\frac{L}{4\pi d^2}

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of  Alpha Centauri A is

\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728

According to the inverse square law for light intensity

\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}

where

\bf I_1= light intensity at distance \bf d_1  

\bf I_2= light intensity at distance \bf d_2  

Let \bf d_2  be the distance we would have to place the 50-watt bulb, then replacing in the formula

\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0
3 years ago
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