2

Korvikt [17]

4+6^2+9•3-10

PEMDAS
P=parentheses
E=exponent
m=multiplication
d=division
a=addition
s=subtraction

*no parentheses in this equation
*exponent = 6^2
6^2=36
4+36+9•3-10
*multiplication = 9•3
9•3=27
4+36+27-10
*addition=4+36+27
4+36+27=67
67-10
*subtraction=67-10
67=10=57

ANSWER=57

please give brainliest itd mean a lot<3

3 0

3 years ago

Travelers who fail to cancel their hotel reservations when they have no intention of showing up are commonly referred to as no-s

notsponge [240]

Answer:

a) 0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) 0 is the most likely value for X.

Step-by-step explanation:

For each traveler who made a reservation, there are only two possible outcomes. Either they show up, or they do not. The probability of a traveler showing up is independent of other travelers. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

No-show rate of 10%.

This means that $p = 0.1$

Four travelers who have made hotel reservations in this study.

This means that $n = 4$

a) What is the probability that at least two of the four selected will turn to be no-shows?

This is $P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486$

$P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036$

$P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001$

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0486 + 0.0036 + 0.0001 = 0.0523$

0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) What is the most likely value for X?

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.6561$

$P(X = 1) = C_{4,1}.(0.1)^{1}.(0.9)^{3} = 0.2916$

$P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486$

$P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036$

$P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001$

X = 0 has the highest probability, which means that 0 is the most likely value for X.

7 0

3 years ago

Please help. Thanks. Work is due tomorrow. Thanks

eimsori [14]

2 hole pizzas equal 16 because 1 pizza has 8 slices so make 1/4 a decimal which is 4 4/16 equals 4 so for 4 friends 1/4 is left and do the same to the 21/3 which equals 7 so 7 +4 = half the amount to fead the friends

7 0

4 years ago

In a sale, a shop reduces all its prices by 10%.

stepladder [879]

Answer:

The phone roughly cost 302 dollars

Step-by-step explanation:

10% of 432=43.2

20% of 432=86.4

432-129.6=302.4

3 0

2 years ago

Answer answer answer answer

Mice21 [21]

Answer:

$12 \leqslant x < 26$

4 0

3 years ago