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lukranit [14]
3 years ago
8

Three eights of the cast in a musical have to sing. What fraction of the cast does not have to sing

Mathematics
2 answers:
lbvjy [14]3 years ago
5 0
5/8 of the cast does not have to sing. You find this by doing 8/8 - 5/8
pychu [463]3 years ago
3 0
3/8 has to sing
8/8 is all
all=haveto+don'thaveto
all=8/8
haveto=3/8
subsitute
8/8=3/8+don'thaveto
subtract 3/8 from both sides
8/8-3/8=5/8
5/8=don'thavetosing

answer is 5/8
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Answer is 37, my friend.

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If the maximum of a data set is 50 and the minimum of the set is 10, what is the range?
Nonamiya [84]

Answer:

40

explanation:

The range is the difference between the smallest and highest numbers in a list or set. To find the range, first put all the numbers in order. Then subtract the lowest number from the highest. The answer gives you the range of the list.

5 0
3 years ago
Reiko takes 2 coins at random from the 3 quarters, 5 dimes, and 2 nickels in her pocket. three coins are tossed. what is the pro
I am Lyosha [343]
The answers are 3/8 and 1/6.

There are 8 possible outcomes in the sample space for flipping 3 coins.  Of those, 3 include two heads and 1 tail.  This makes the probability 3/8.

Choosing a quarter for the first coin is 3/10, since there are 3 quarters out of 10 coins.  After this, choosing a dime is given by 5/9, since there are 5 dimes out of 9 coins left:
3/10(5/9) = 15/90 = 1/6
5 0
3 years ago
1. Trapezoid KLMN has vertices K(1, 3) , L(3, 1) , M(3, 0) , and N(1, −2) .
chubhunter [2.5K]

Answer:


Step-by-step explanation:

(A) The vertices of the trapezoid KLMN are K(1, 3) , L(3, 1) , M(3, 0) , and N(1, −2)

Now, KL= \sqrt{(3-1)^{2}+(1-3)^{2}}=\sqrt{8}[/tex]

LM=\sqrt{(3-3)^{2}+(0-1)^{2}}=1

MN=\sqrt{(1-3)^{2}+(-2-0)^{2}}=\sqrt{8}

NK=\sqrt{(1-1)^{2}+(-2-3)^{2}}=5

Now, as KL and MN  are equal, therefore, KLMN is an isosceles trapezoid.

(B) Since  m∠XWY=47°, therefore ∠XWZ=47+47=94° and ∠ZYW=18°, therefore ∠XYZ=36°( as diagonals bisect the angles)

In ΔXWO,

∠XWO+∠WXO+∠WOX=180°(Angle sum property)

∠WXO=43°

Also, from ΔXOY,

∠OXY=72° using the angle sum property.

Therefore, ∠WXY=43+72=115°

Now, sum of all the angles of a quadrilateral is equal to 360°, therefore

∠WXY+∠XYZ+∠YZW+∠ZWX=360°

115°+36°+∠YZW+94°=360°

∠YZW=115°

Therefore,∠WZY=115°

(C) Since, AB and CD are the two parallel lines as the slope of both the sides are equal.

LetM be the mid point of AB, therefore M=(\frac{-2+4}{2}, \frac{4+3}{2})

=(1,\frac{7}{2})

Also, let N be the mid point of DC, therefore,

N=(\frac{4-2}{2},\frac{-2-5}{2})

=(1,\frac{-7}{2})

Now, length of the mid segment MN= \sqrt{(1-1)^{2}+(\frac{7}{2}+\frac{7}{2})^{2}}=\sqrt{0+49}=7 cm

(D)Given: Kite PQRS,  TS=6cm and TP=8cm

Then, From triangle TSP, we have

(SP)^{2}=(TS)^{2}+(TP)^{2}

(SP)^{2}=36+64

SP=10cm

8 0
3 years ago
I need help! ASAP!:(
kirill [66]

Answer:

18(40/360) + 9( 140/360) -3(90/360) -7(90/360)

=2 + 3.5 - 0.75 - 1.75

= $3

4 0
3 years ago
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