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N76 [4]
2 years ago
15

6.339m plus 0.170m plus 30.4m

Mathematics
1 answer:
77julia77 [94]2 years ago
7 0

Answer:

36.909m

Step-by-step explanation:

6.339+0.170+30.4

=36.909m

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Simplify 10 divide 5 + 3 x 2.
mylen [45]

Answer:

d 8

Step-by-step explanation:

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2 years ago
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GRAHAM CORPORATION
astra-53 [7]

Answer:

<h2>Net proceed= s-0.98nd-11</h2>

Step-by-step explanation:

Given that the shares cost $d

n shares will be =$dn

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sold the share for $s

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collect like terms

Net proceed= s-(nd-0.02nd)-11

Net proceed= s-0.98nd-11

5 0
3 years ago
ms anderson bought 45 baseball hats. Mets hats cost $18 and Yankee hats cost $24. If she spent $1050, how many of each did she b
DerKrebs [107]
First, you have to set up two equations. X + y = 45 18x + 24y = 1,050 To do this, you have to solve for an exponent by cancelling out the other one. To solve for y, you multiply everything in the first equation by -18. -18x -18y = -810 This cancels out the x and leaves you with 6y = 240. Divide both sides by 6 and you have the number of Yankee hats that were bought. Then, you have to substitute 40 for y in the first equation and solve for x.
5 0
3 years ago
X/3 +17 less than or equal to 15
Sliva [168]

x \leqslant  - 6

x/3+17is less than or equal to 15

x/3is less than or equal to 15-17

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5 0
3 years ago
Assuming boys and girls are equally​ likely, find the probability of a couple having a baby girl when their sixth child is​ born
Ivanshal [37]

Answer:  The required probability of having 6th girl is 0.5.

Step-by-step explanation:  Given that boys and girls are equally likely.

We are to find the probability of a couple having a baby girl when their sixth child is​ born, given that the first five children were all girls.

Since the events of having a boy and a girl are independent of each other, so

the probability of having 6th girl dose not depend on the birth of the first five girls.

We know that there are only two possible cases (either a boy or girl will born).

So, sample space, S = {G, B}  and the event E of having a girl is, E = {G}.

That is, n(S) = 2 and n(E) = 1.

Therefore, the probability of event E is given by

P(E)=\dfrac{n(E)}{n(S)}=\dfrac{1}{2}=0.5.

Thus, the required probability of having 6th girl is 0.5.

3 0
3 years ago
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