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Airida [17]
3 years ago
7

4/9 of the students are in one class are boys. What fraction of the class re girls

Mathematics
1 answer:
kkurt [141]3 years ago
5 0
9/9 ( 1 whole class) - 4/9 (number of boys) = 5/9.
5/9 of the class are girls. I hope hope this helps is helps you!
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The mean of the masses of five articles is 215 g. When another article is added, the mean of the masses of the six articles is 2
lbvjy [14]

Mass of the sixth article will be :

\boxed{ \boxed{245 \: g}}

\mathrm{✌TeeNForeveR✌}

4 0
3 years ago
Find the missing probability. P(B)=720,P(A|B)=14 ,P(A∩B)=?
Ivan

Answer:

7/80

Step-by-step explanation:

Given that: P(B) = 7 / 20, P(A|B)= 1 / 4

Bayes theorem is used to mathematically represent the conditional probability of an event A given B. According to Bayes theorem:

P(A|B)=\frac{P(A \cap B)}{P(B)}

Where P(B) is the probability of event B occurring, P(A ∩ B) is the probability of event A and event B occurring and P(A|B) is the probability of event A occurring given event B.

P(A|B)=\frac{P(A \cap B)}{P(B)}\\\\P(A \cap B)=P(A|B)*P(B)\\\\Substituting:\\\\P(A \cap B)=1/4*7/20=7/80\\\\P(A \cap B)=7/80

6 0
3 years ago
Which relation is a function?
Veseljchak [2.6K]

Answer: B


Step-by-step explanation:

because in A it says 2=3 and 2=2 that is not a function each number has to have an outcome of one possible answer if that is true it is a function.

7 0
3 years ago
A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome
Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), <em>p</em> = 0.52

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.

(1)

Compute the probability of overbooking  as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

                                        =P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

=P(X

Thus, the probability that the flight has empty seats is 0.4625.

4 0
3 years ago
Renee is in charge of the school carnival for 380 students. She has 47 boxes of prizes. Each box has 22 prizes. She wants to mak
rodikova [14]
Multiply 47x22 and you'll get a total of 1034 prizes
5 0
3 years ago
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