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enot [183]
2 years ago
5

Help Me please it’s mathhh

Mathematics
1 answer:
enyata [817]2 years ago
3 0

Answer: t = 5

4 = 2 + 2/ 5 t

4 = 2 +2/ 5

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Susan used 1/2 gallons of gas on Sunday and 3 5/8 gallons of gas on Monday. How many more gallons of gas did she use on Monday?
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Answer: 3 and 1/8 more

Step-by-step explanation:

You can start by converting to the same fraction and making the second one improper, so 29/8 and 4/8. Subtract them to find the difference and  you are left with 25/8 or 3 1/8 gallons.

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3 years ago
What is the solution to the equation 2 (x-3) ^2 =13 help help ASAP !
olganol [36]

Answer:

3+sqrt(13/2)

3-sqrt(13/2)

Step-by-step explanation:

2 (x-3) ^2 =13

Divide each side by 2

2/2 (x-3) ^2 =13/2

(x-3) ^2 =13/2

Take the square root of each side

sqrt((x-3) ^2) =±sqrt(13/2)

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4 0
3 years ago
Read 2 more answers
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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