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GarryVolchara [31]
3 years ago
6

25 take away 2 / 9 ​

Mathematics
1 answer:
Serga [27]3 years ago
4 0

Answer:

24.7 repeated

Step-by-step explanation:

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There are 140 members in a community sports program. Sixty-five percent of these members play soccer.
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75 people do not play soccer
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Help pls<br> 4 1/5 of 8 2/10 is what number ?
soldi70 [24.7K]

Answer:

34.44 or 34 11/25

Step-by-step explanation:

4 1/5 ⋅ 8 2/10 = ?

21/5 ⋅ 82/10 = ?

21/5 ⋅ 41/5 = 861/25

861/25 = 34 11/25 or 34.44

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3 years ago
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5. Simplify: 56 = 7 - 3 [4 + {8 - 4 ( 4 + 5 - 3 )}]​
shutvik [7]

Answer:

36

Step-by-step explanation:

56=7-3[4+{8-4(4+5-3)}]

56=7-3[4+{4(4+5-3)}]

56=7-3[4+{16+20-12)}]

56=7-3[4+24]

56=4[28]

56=92

56-56=92-56

36

3 0
3 years ago
A members-only speaker series allows people to join for $28 and then pay $4 for every event attended. What is the total cost for
wariber [46]

Given:

Joining fee = $28

Fee of each event = $4

To find:

Total cost for someone to attend 4 events.

Solution:

Let the number of events be x and total fee be y.

Fee for 1 event = $4

Fee for x events = $4x

Joining fee remains constant. So, the total fee is

y=28+4x

Substitute x=4 in this equation.

y=28+4(4)

y=28+16

y=44

Therefore, total cost of 4 events is $44.

8 0
3 years ago
A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
3 years ago
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