Wich pair of variables shows a clear casual link

Last month haos dog ate 40 cans of dog food in thirty days how many cans should hao buy to feed his dog for the next six days

Tom [10]

30 Days = 40 Cans
6 Days (30/5 Days) = 8 Cans (40/5 Cans)

7 0

3 years ago

3 thirtieth simplified to

Sedbober [7]

1/10. You can do this by dividing both numbers by three. Hope this helped!

5 0

4 years ago

after making 20 servings of pasta a chef has used 30 cloves of garlic the shaft use 6 cloves to make the first four servings how

Rus_ich [418]

15 gloves were used for 10 servings

8 0

4 years ago

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

Two mechanics worked on a car. The first mechanic charged S95 per hour, and the second mechanic charged $115 per hour.

natima [27]

Answer:

First mechanic worked for 10 hours and second mechanic worked for 15 hours.

Step-by-step explanation:

Let first mechanic worked for X hours and second mechanic worked for Y hours .

Now, according to the question , they worked for a total of 25 hours .

Thus ,

X + Y = 25.

Y = 25 - X -(1)

Also, first mechanic is charging $ 95 per hour and second mechanic is charging $115 per hour .

Thus, total money charged by them will be 95X + 115Y .

Which is given to be $2675 in the question.

Thus 95X + 115Y = 2675. -(2)

Putting equation 1 in equation 2 we get ,

95X + 2875 - 115X = 2675

20X = 200

X = 10 hours.

Thus , Y = 25 - X = 15 hours.

Thus first mechanic worked for 10 hours and second mechanic worked for 15 hours.

5 0

3 years ago