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earnstyle [38]
3 years ago
9

Find the area of the parallelogram shown.

Mathematics
2 answers:
faust18 [17]3 years ago
6 0
Can you put the photo
Ann [662]3 years ago
5 0
Where is the parallelogram?
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emily bought 3 vases of flowers.the most expensive vase of flowers was 12.00 and least expensive was 8.00 which is a reasonable
iogann1982 [59]

Answer:$30

Step-by-step explanation:8+12=20 and I would figure she had another one with a price inbetween the two so I chose 10 and added 20+10=30

4 0
3 years ago
88.83 is what percent of 21?
ozzi
423%. Calculate this with \frac{88.83}{21}
5 0
4 years ago
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Y=300-5x in standard form
Akimi4 [234]
<span>y = 300 - 5x

Standard form:
ax + by = c

</span>y = 300 - 5x
<span>5x + y = 300

Then to find the y-intercept, insert a 0 into the 'x'.
5(0) + y = 300
y = 300
The y-intercept is 300.

To find the x-intercept, insert a 0 into the 'y'.
5x + (0) = 300
5x = 300
x = 60

The x-intercept is 60.</span>
8 0
3 years ago
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A set of numbers satisfies Benford’s Law if the probability of a number starting with digit d is P(d) = log(d + 1) – log(d).
irakobra [83]

Interpreting the graph and the situation, it is found that the values of d that can be included in the solution set are 1 and 4.

----------------------

  • According to Benford's law, the probability of a number starting with digit is d is:

P(d) = \log{(d + 1}} - \log{d}

  • A number can start with 10 possible digits, ranging from 1 to 9, which are all integer digits.
  • Thus, d can only assume integer digits.
  • In the graph, the solution is d < 5.
  • The integer options for values of d are 1 and 4.
  • For the other options that are less than 5, they are not integers, so d cannot assume those values.

A similar problem is given at brainly.com/question/16764162

8 0
2 years ago
Read 2 more answers
The top and bottom margins of a poster are each 15 cm and the side margins are each 10 cm. If the area of printed material on th
12345 [234]

Answer:

the dimension of the poster = 90 cm length and 60 cm  width i.e 90 cm by 60 cm.

Step-by-step explanation:

From the given question.

Let p be the length of the of the printed material

Let q be the width of the of the printed material

Therefore pq = 2400 cm ²

q = \dfrac{2400 \ cm^2}{p}

To find the dimensions of the poster; we have:

the length of the poster to be p+30 and the width to be \dfrac{2400 \ cm^2}{p} + 20

The area of the printed material can now be:  A = (p+30)(\dfrac{2400 }{p} + 20)

=2400 +20 p +\dfrac{72000}{p}+600

Let differentiate with respect to p; we have

\dfrac{dA}{dp}= 20 - \dfrac{72000}{p^3}

Also;

\dfrac{d^2A}{dp^2}= \dfrac{144000}{p^3}

For the smallest area \dfrac{dA}{dp }=0

20 - \dfrac{72000}{p^2}=0

p^2 = \dfrac{72000}{20}

p² = 3600

p =√3600

p = 60

Since p = 60 ; replace p = 60 in the expression  q = \dfrac{2400 \ cm^2}{p}   to solve for q;

q = \dfrac{2400 \ cm^2}{p}

q = \dfrac{2400 \ cm^2}{60}

q = 40

Thus; the printed material has the length of 60 cm and the width of 40cm

the length of the poster = p+30 = 60 +30 = 90 cm

the width of the poster = \dfrac{2400 \ cm^2}{p} + 20 = \dfrac{2400 \ cm^2}{60} + 20  = 40 + 20 = 60

Hence; the dimension of the poster = 90 cm length and 60 cm  width i.e 90 cm by 60 cm.

4 0
3 years ago
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