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ruslelena [56]
3 years ago
11

1. Covert 4 g to mg.

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
6 0

Answer:

1. 4 to 4,000mg

2. 14 to 0.14m

3. 450 to 0.45L

4. 1.5 to 150cm

5. 0.2 to 200mL

You might be interested in
We learned in that about 69.7% of 18-20 year olds consumed alcoholic beverages in 2008. We now consider a random sample of fifty
maw [93]

Answer:

(1) The expected number of people who would have consumed alcoholic beverages is 34.9.

(2) The standard deviation of people who would have consumed alcoholic beverages is 10.56.

(3) It is surprising that there were 45 or more people who have consumed alcoholic beverages.

Step-by-step explanation:

Let <em>X</em> = number of adults between 18 to 20 years consumed alcoholic beverages in 2008.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.697.

A random sample of <em>n</em> = 50 adults in the age group 18 - 20 years is selected.

An adult, in the age group 18 - 20 years, consuming alcohol is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 50 and <em>p</em> = 0.697.

The probability mass function of a Binomial random variable <em>X</em> is:

P(X=x)={50\choose x}0.697^{x}(1-0.697)^{50-x};\ x=0,1,2,3...

(1)

Compute the expected value of <em>X</em> as follows:

E(X)=np\\=50\times 0.697\\=34.85\\\approx34.9

Thus, the expected number of people who would have consumed alcoholic beverages is 34.9.

(2)

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{np(1-p)}=\sqrt{50\times 0.697\times (1-0.697)}=10.55955\approx10.56

Thus, the standard deviation of people who would have consumed alcoholic beverages is 10.56.

(3)

Compute the probability of <em>X</em> ≥ 45 as follows:

P (<em>X</em> ≥ 45) = P (X = 45) + P (X = 46) + ... + P (X = 50)

                =\sum\limits^{50}_{x=45} {50\choose x}0.697^{x}(1-0.697)^{50-x}\\=0.0005+0.0001+0.00002+0.000003+0+0\\=0.000623\\\approx0.0006

The probability that 45 or more have consumed alcoholic beverages is 0.0006.

An unusual or surprising event is an event that has a very low probability of success, i.e. <em>p</em> < 0.05.

The probability of 45 or more have consumed alcoholic beverages is 0.0006. This probability value is very small.

Thus, it is surprising that there were 45 or more people who have consumed alcoholic beverages.

6 0
3 years ago
Can someone. help me solve this pls .
olchik [2.2K]

Answer:

1.108

729 =  {n}^{6} \\  \sqrt[6]{729}  =   \sqrt[6]{ {n}^{6} }  \\ n = 1.108

7 0
3 years ago
Question is in picture. Please show your work, so I can see how you got the answer.
weqwewe [10]
It’s D! they have one intersect
5 0
3 years ago
Read 2 more answers
Each student in a class plays one of three sports soccer, volleyball or basketball. 1/4 of them play soccer. 3/5 of them play vo
Setler79 [48]

Answer: 3/20 of the class

Step-by-step explanation:

Add the amount of students that play soccer to amount of students that play volleyball

1/4 + 3/5 = 17/20

Subtract 17/20 (soccer and volley ball) from 20/20 or 1

20/20 - 17/20 = 3/20

7 0
3 years ago
2. There are 1500 candies in a candy dish. Each day half of the candy is taken out.
katen-ka-za [31]

Answer:

so easy and obvious dude

Step-by-step explanation:

5 0
3 years ago
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