Ask question

Ask question

All categories

2 years ago

5

A wedding planner she did the 234 gas at a reception so that each table have the same number of guest if each table could see no

more than 20 people what is the largest possible number of guest at each table

1 answer:

Studentka2010 [4]2 years ago

6 0

Answer:

the largest number of people she can sit at a table is 18.

Step-by-step explanation:

there would be 13 tables

You might be interested in

In a class of 30 students, 6 have a brother and 12 have a sister. There are 14 students

Elena-2011 [213]

30 students

14 have no siblings

that leaves 16 with siblings

6 have brothers and 12 have sisters which = 18 so 2 of the students have both a brother and a sister

2/30 students have a brother and a sister

1/15 chance

3 0

2 years ago

If 12 > -15, which of the following must be true?

Tcecarenko [31]

Answer:

12 is to the right of -15 on a number line.

Step-by-step explanation:

Hope this helps! - CJ

5 0

2 years ago

Rewrite as a simplified fraction.

Sauron [17]

Answer:

0.2 = 2/9

Step-by-step explanation:

Hope my answer has helped you!

5 0

2 years ago

Help meeee pleaseeeeee

Leto [7]

Answer: B, angle 1 and angle 7.

Step-by-step explanation: Of all the options provided, B is the only one where the the two angles are not congruent to each other. Also, same side interior angles are congruent, so...

5 0

2 years ago

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development

Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

5 0

2 years ago