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andre [41]
2 years ago
11

of the students in ms. hillers class, 27% have blonde hair, 0.33 have black hair and 2/5 have brown hair. which hair color is mo

st common?
Mathematics
1 answer:
aleksandrvk [35]2 years ago
6 0

Answer:

Brown hair is the most common

Step-by-step explanation: 0.33 as a percent is 33% and 2/5 as a percent is 40% and 40 is the most the 40% wins

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PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!
zalisa [80]

Answer:

The answe is 2/5m=4

Step-by-step explanation:

If he ate 4 of them, but it means he ate 2/5 muffins,

4 muffins = 2/5 muffins

You substitute x muffins for how much muffins she baked in all -

2/5 * x muffins = 4  or 2/5m = 4

Hope that helped :) <3

5 0
2 years ago
For an activity in class, student used pieces of string to measure road on a map. They used a map scale to determine the distanc
ivanzaharov [21]

Answer:

Step-by-step explanation:

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7 0
3 years ago
Work out 30 – 3 + 7 x 2
strojnjashka [21]
The correct answer is 41
7 0
2 years ago
Read 2 more answers
If f x equals x -3 + g x equals 1 - x 2 - 9 find ( f og) (x and give any restrictions on its domain
evablogger [386]
\begin{gathered} f(x)=x-3 \\ g(x)=\frac{1}{x^2-9} \\ (\text{fog)(x)}=\frac{1}{x^2-9}-3 \end{gathered}

the denominator cannot be zero, because the division by zero is not defined, therefore:

\begin{gathered} x^2-9=0 \\ \text{Solving for x:} \\ x^2=9 \\ \sqrt[]{x^2}=\sqrt[]{9} \\ x=\pm3 \end{gathered}

Therefore the domain of (f o g)(x) is:

D\colon(-\infty,-3)\cup(-3,3)\cup(3,\infty)

3 0
1 year ago
Two isosceles triangles share the same base. Prove that the medians to this base are collinear. (There are two cases in this pro
Anna71 [15]

Answer:

Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

→Case 1.  When vertices A and P are opposite side of Base BC.

In Δ ABD and Δ ACD

AB= AC   [Given]

AD is common.

BD=DC  [median of a triangle divides the side in two equal parts]

Δ ABD ≅Δ ACD [SSS]

∠1=∠2 [CPCT].........................(1)

Similarly, Δ PBD ≅ Δ PCD [By SSS]

 ∠ 3 = ∠4 [CPCT].................(2)

But,  ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

2 ∠2 + 2∠ 4=360° [using (1) and (2)]

∠2 + ∠ 4=180°

But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

In ΔPBD and ΔPCD

PB=PC[given]

PD is common.

BD =DC [Median of a triangle divides the side in two equal parts]

ΔPBD ≅ ΔPCD  [SSS]

∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

∴ P,A,D lie on a line i.e they are collinear.




3 0
3 years ago
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