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olga_2 [115]
4 years ago
7

Solve this ratio 18 hours to 2 days as a fraction in simplest form

Mathematics
1 answer:
dlinn [17]4 years ago
8 0
Answer 3/8. 24 hours in a day and there’s 2 days which is 48 hours. 18/48 simplifies to 3/8
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Answer which of the following is not a solution
masya89 [10]

Answer:

I'm assuming (4,2) since its not gathered with the other points

8 0
3 years ago
calculate the volume of each cone. Use 3.14 for π. Round answers to the nearest hundredth if necessary.​
coldgirl [10]

Answer:

2355 cm^2

Step-by-step explanation:

The volume of a cone is given by

V = 1/3 pi r^2 h

V = 1/3 (3.14) (15)^2 (10)

V =2355

8 0
3 years ago
When solving quadratic equation, what do the X value stand for as it relates to plotting the graph of the equation?
Ber [7]

Step-by-step explanation:

step 1. The x values refer to the domain of the function.

step 2. if you set y = 0 and solve the quadratic equation the x values refer to the point the graph crosses the x axis.

7 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
What is the value of z?
Amiraneli [1.4K]

Answer:

if a z-score is equal to +1, it is 1 standard deviation above the mean.

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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