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VMariaS [17]
3 years ago
7

Can someone help me with this math question 2^x=2^3

Mathematics
1 answer:
Soloha48 [4]3 years ago
7 0
2ˣ = 2³

Since 2ˣ is EQUAL to  2³  and since 2 is equal to 2, then the solution is a must that the 2 exponents are equal.
Then x = 3
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Thx for helping me. cljck on pic
Whitepunk [10]

Answer: 56


Step-by-step explanation: 100 multiplied  by 50% is 50 + 12% =56


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4 years ago
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$3100 boat, 5.5% sales tax
Serjik [45]

Answer:

3100.00 * 5.5% (170.50) = 3270.50

Step-by-step explanation:

3100.00 * 5.5% (170.50) = 3270.50

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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Liula [17]

Answer:

They are equivalent

Step-by-step explanation:

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6 0
4 years ago
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2 + 2 What is the Answer ?
kifflom [539]

Answer:

The answer is 4!

Step-by-step explanation:

2 + 2 can be rewritten as 1+1+1+1

both equal 4

5 0
3 years ago
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