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mafiozo [28]
3 years ago
10

What is 140% of 6000

Mathematics
2 answers:
ludmilkaskok [199]3 years ago
7 0

The answer would be 8400 Hope I helped

kolbaska11 [484]3 years ago
5 0

Answer:

8400

Step-by-step explanation:

6000 x 1.40 = 8400

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The way you wrote it is how you write it unless you want to do scientific notation then the answer would be, 5x10^30 because for it to be in scientific notation it has to be between 10 and 1 so I moved the decimal spot over 30 to get it into scientific notation. Hope this helps out.
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The expression (y to the power of 20)(y to the power of −5)2 is equivalent to yn. What is the value of n?
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The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medic
Ksivusya [100]

Answer:

(a) P (X = 0) = 0.006

(b) P (X = 1) = 0.0403

(c) P (X ≥ 2) = 0.9537

(d) P (X > 5) = 0.1664

Step-by-step explanation:

Let <em>X</em> = number of successful appeals.

The probability that an appeal is successful is, <em>p</em> = 0.40.

The sample size is, <em>n</em> = 10.

The random variable X\sim Bin(n=10, p=0.40).

The probability function of a Binomial distribution is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x}

(a)

The probability that none of the appeals is successful is:

P(X=0)={10\choose 0}(0.40)^{0}(1-0.40)^{10-0}\\=1\times1\times0.0060466176\\=0.0060466176\\\approx0.006

Thus, the probability that none of the appeals is successful is 0.006.

(b)

The probability that exactly one of the appeals will be successful is:

P(X=1)={10\choose 1}(0.40)^{1}(1-0.40)^{10-1}\\=10\times0.40\times0.010077696\\=0.040310784\\\approx0.0403

Thus, the probability that exactly one of the appeals will be successful is 0.0403.

(c)

The probability that at least two of the appeals will be successful is:

P(X\geq 2)=1-P(X

Thus, the probability that at least two of the appeals will be successful is 0.9537.

(d)

The probability that more than half of the appeals will be successful is:

P(X> 5)=1-P(X\leq 5)\\=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)+P(X=5)\\=1-[{10\choose 0}(0.40)^{0}(1-0.40)^{10-0}]-[{10\choose 1}(0.40)^{1}(1-0.40)^{10-1}]\\-[{10\choose 2}(0.40)^{2}(1-0.40)^{10-2}]-[{10\choose 3}(0.40)^{3}(1-0.40)^{10-3}]\\-[{10\choose 4}(0.40)^{4}(1-0.40)^{10-4}]-[{10\choose 5}(0.40)^{5}(1-0.40)^{10-5}]\\=1 - 0.006-0.0403-0.1209-0.2150-0.2508-0.2006\\=0.1664

Thus, the probability that more than half of the appeals will be successful is 0.1664.

7 0
3 years ago
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