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Tomtit [17]
3 years ago
12

Daria bought two same sized poster boards. she cut the posterboards into equal sized pieces to make placemats for her dinner ges

ts. she cut the first posterboard into 5 pieces and the second posterboard into 2 pieces. she will continue to cut the pieces of posterboard so that each one is divided into the same number of equal sized pieces. what is the least number of equal sized pieces each posterboard could have?
Mathematics
1 answer:
Gelneren [198K]3 years ago
4 0
<span>The least number of equal sized pieces each posterboard could have is given by the lowest common multiple of 5 and 2.

The LCM of 5 and 2 is 10.

Therefore, the </span><span>least number of equal sized pieces each posterboard could have is 10.</span>
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Find the horizontal distance between points C(3.5, −0.25) and D(0.5, −0.25).
ANEK [815]

Answer:

3 units

Step-by-step explanation:

Since the y-coordinate is the same. Therefore, there is a straight horizontal line between the two points. So, again, since the y-coordinates are the same, we can subtract C(x) (the x-coordinate for point C) from D(x) (the x-coordinate for point C)

C(x) = 3.5

D(x) = 0.5

3.5 - 0.5 = 3

Therefore, the distance between the two points is 3 units

7 0
3 years ago
Explain for brainliest (pog)
ElenaW [278]

Answer:

b. exactly one solution

Step-by-step explanation:

-5(z+1)=-2z+10

(first, distribute the 5 into the parenthesis)

-5z-5=-2z+10

(next, move the variables to the left and constants to the right)

-5z+2z=10+5

(combine like terms)

-3z=15

(isolate to variable to find the solution)

z=-5

i hope this helps :)

5 0
3 years ago
Read 2 more answers
Yahoo creates a test to classify emails as spam or not spam based on the contained words. This test accurately identifies spam (
Amiraneli [1.4K]

Answer and Step-by-step explanation:

The computation is shown below:

Let us assume that

Spam Email be S

And, test spam positive be T

Given that

P(S) = 0.3

P(\frac{T}{S}) = 0.95

P(\frac{T}{S^c}) = 0.05

Now based on the above information, the probabilities are as follows

i. P(Spam Email) is

= P(S)

= 0.3

P(S^c) =  1 - P(S)

= 1 - 0.3

= 0.7

ii. P(\frac{S}{T}) = \frac{P(S\cap\ T}{P(T)}

= \frac{P(\frac{T}{S}) . P(S) }{P(\frac{T}{S}) . P(S) + P(\frac{T}{S^c}) . P(S^c) }

= \frac{0.95 \times 0.3}{0.95 \times 0.3 + 0.05 \times 0.7}

= 0.8906

iii. P(\frac{S}{T^c}) = \frac{P(S\cap\ T^c}{P(T^c)}

= \frac{P(\frac{T^c}{S}) . P(S) }{P(\frac{T^c}{S}) . P(S) + P(\frac{T^c}{S^c}) . P(S^c) }

= \frac{(1 - 0.95)\times 0.3}{ (1 -0.95)0.95 \times 0.3 + (1 - 0.05) \times 0.7}

= 0.0221

We simply applied the above formulas so that the each part could come

8 0
3 years ago
In a biathlon , jackson cycled 9 miles in 3/4 of an hour in the first event. He than ran, on average, 7 miles per hour slower in
lozanna [386]

Answer:

7.5 miles

Step-by-step explanation:

Given

  • In cycling,He covered 9 miles in \frac{3}{4} hours,Therefore his avg cycling speed, u=\frac{Distance}{Time} =\frac{9}{\frac{3}{4} } =12\frac{miles}{hour}
  • In second event, It took him 1\frac{1}{2} =\frac{3}{2}hours to complete,And his avg speed,v is  7\frac{miles}{hour} slower than in previous event

⇒v= u-7

v= 12-7

v=5

Distance covered in 1st event=9miles

Distance covered in 2nd event=Speed\times Time= 5\times \frac{3}{2} =7.5miles

4 0
3 years ago
A miniature golf course recently provided its customers with 6 golf balls, of which 3 were blue. What is the experimental probab
Schach [20]
The next probable answer is 1:2
5 0
3 years ago
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