50 ÷ 5 (5 – 3) + 5 ÷ 1/5

What percent of his overall bill did beau save?

Katen [24]

33.3333333333333333333333333333333333333% um, you get the point right? 205-143.50 = 61.50

205/61.50 = 3.333333333333333333333333, or 33.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333%

5 0

3 years ago

Read 2 more answers

Amir can travel 121.72 miles on 6.8 gallons of gas.

ASHA 777 [7]

The unit consumption of gas can be obtained by dividing the total distance traveled in miles by the total gallons of gas consumed by the trip. This is shown below:

121.72 miles / 6.8 gallons = 17.9

Therefore, on a gallon of gas, Amir can travel 17.9 miles, which is letter B among the choices.

4 0

3 years ago

Please help me out ~ 20 points

nirvana33 [79]

Answer:

12.5 sq. units

Step-by-step explanation:

Area of a Trapezoid Formula: (see the image attached)

7.5 + 2.5 = 10

10/2 = 5

5(2.5) = 12.5

So, the area of the trapezoid is 12.5 sq. units.

8 0

3 years ago

Enter your answer in the box to solve the problem. 4/6=?/12

ser-zykov [4K]

Answer:

4/6=8/12

Step-by-step explanation:

If the denominator is multiplied by 2, the numerator needs to do the same to keep the fraction equal.

6*2=12

4*2=8

4/6=8/12

4 0

3 years ago

Read 2 more answers

The probability that a single radar station will detect an enemy plane is 0.65.

taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

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 combinatios 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.

In this problem, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that $n = 0.65$ .

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which $P(X = 0) \leq 0.02$ .

n = 1.

$P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35$

n = 2

$P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225$

n = 3

$P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429$

n = 4

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

$E(x) = np$

So when $n = 7$

$E(x) = 7*(0.65) = 4.55$

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

6 0

3 years ago