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satela [25.4K]
4 years ago
7

11% as a decimal= 200% as a decimal= 20.07% as a decimal 198.4% as a decimal

Mathematics
2 answers:
Hitman42 [59]4 years ago
8 0
To convert from a percentage to a decimal, move the decimal twice to the left

11% = 0.11
200% = 2.00
20.07% = 0.2007
198.4% = 1.984

Hope this helps!
larisa [96]4 years ago
3 0

Answer:

11% = .11

200%= 2

20.07%= .2007

198.4%= 1.984

Step-by-step explanation:

Whenever you convert percent to decimal you have to divide the percent by 100.

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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
How would i solve for X? please explain.
kap26 [50]
In problems like this the outer and inner part are in equal ratio
4 0
3 years ago
Y−3=2(x+1) <br><br> Find The Value of -1
murzikaleks [220]

Answer:

3

Step-by-step explanation:

I'm assuming you're saying the value of x is -1?

y-3=2(-1+1)

y-3=2(0)

y-3=0

y=3

3 0
3 years ago
Compute a 99% confidence interval for the mean time between births at this hospital. 1. What is the lower limit of this confiden
MrMuchimi

Answer:

90% confidence interval between 118.64 ounces and 124.16 ounces

99% confidence interval between 117.13 ounces and 125.67 ounces

Step-by-step explanation:

8 0
4 years ago
What is the area of the following parallelogram in square centimeters?
Llana [10]

Answer:

A. 60 cm 2

Step-by-step explanation:

hope this helps

A=base height=30·2=60

6 0
3 years ago
Read 2 more answers
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