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3 years ago

What is the diameter, in meters of the circular area that gets watered by the sprinkler?

Mathematics

2 answers:

natka813 [3]3 years ago

5 0

Answer:

The answer is 12 m

Step-by-step explanation: As we know, the relationship between the diameter and radius, is that it is 2 times the amount of it. Now in your particular equation, 36 is radius^2. So all we have to do, is take the square root of 36, and then multiply that number by 2, to find the diameter. The square root of 36 is 6, so 6 x 2= 12 m, which is the diameter.

Please let me know if you have any questions, thank you!

Schach [20]3 years ago

3 0

The answer is 12 m.

Because you're not studying!

Next time do your mom.

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Use the definition of absolute value to write f as a piecewise-defined function

ycow [4]

we are given

any absolute function

Let's assume that function as

f(x)=|x|

now, we can write it in piecwise form

we know that value of f(x) will always be positive

so, we write function such that it results in positive value

so, we can write as

$f(x)=-x;....x\leq 0$

$f(x)=x;....x>0$

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3 years ago

Pleaseeeee helppppDetermine how many actual fish food boxes fit in the shipping box. Show your work. Find the volume of the ship

Alex Ar [27]

Shipping box is a cube, with units

Fish food is an unitary cube, that means have x,y,z measures the same

Shipping box is a parallelepip box. Rectangular

Volume of shipping box is found by formula

V = X•Y•Z

X = 3ft, Y=3 3/4 ft, Z=3 1/4 ft

then V = (3•3+ 3•3/4)•(3+1/4) = (11 +1/4)• (3+1/4) = 33 + 11/4 + 3/4 + 1/16

V = 33 + 2 + 3/4+ 3/4 +1/16 = 35 + 25/16 = 34+ 9/16

Now, find volume using packing cubes

How many fish food boxes fit

In X direction it fits 3/(1/4) = 12

In Y direction it fits (3+3/4)/(1/4)= 15

In Z direction it fits (3+1/4)/(1/4) = 10

Now then multiply 12x15x10= 1800 fish boxes fit in larger box

3 0

2 years ago

According to an airline, flights on a certain route are on time 90 90% of the time. Suppose 20 20 flights are randomly selecte

Brilliant_brown [7]

Answer:

(a) Yes, the above experiment is a binomial distribution.

(b) Probability that exactly 18 flights are on time is 28.5%.

(c) Probability that at least 18 flights are on time is 67.7%.

(d) Probability that fewer than 18 flights are on time is 32.3%.

(e) Probability that between 17 and 19 flights, inclusive, are on time is 74.5%.

Step-by-step explanation:

We are given that according to an airline, flights on a certain route are on time 90% of the time.

Suppose 20 20 flights are randomly selected and the number of on time flights is recorded.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

where, n = number trials (samples) taken = 20 flights

r = number of success

p = probability of success which in our question is probability that

flights on a certain route are on time, i.e; p = 0.90

Let X = Number of flights on a certain route that are on time

So, X ~ Binom(n = 20, p = 0.90)

(a) Yes, the above experiment is a binomial distribution as the probability of success is the probability in a single trial.

And also, each flight is independent of another.

(b) Probability that exactly 18 flights are on time is given by = P(X = 18)

P(X = 18) = $\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}$

= $190 \times 0.90^{18} \times 0.10^{2}$

= 0.285

Therefore, probability that exactly 18 flights are on time is 28.5%.

(c) Probability that at least 18 flights are on time is given by = P(X $\geq$ 18)

P(X $\geq$ 18) = P(X = 18) + P(X = 19) + P(X = 20)

= $\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}+\binom{20}{19} \times 0.90^{19} \times (1-0.90)^{20-19}+\binom{20}{20} \times 0.90^{20} \times (1-0.90)^{20-20}$

= $190 \times 0.90^{18} \times 0.10^{2}+20 \times 0.90^{19} \times 0.10^{1}+1 \times 0.90^{20} \times 0.10^{0}$

= 0.677

Therefore, probability that at least 18 flights are on time is 67.7%.

(d) Probability that fewer than 18 flights are on time is given by = P(X<18)

P(X < 18) = 1 - P(X $\geq$ 18)

= 1 - 0.677 = 0.323

Therefore, probability that fewer than 18 flights are on time is 32.3%.

(e) Probability that between 17 and 19 flights, inclusive, are on time is given by = P(17 $\leq$ X $\leq$ 19)

P(17 $\leq$ X $\leq$ 19) = P(X = 17) + P(X = 18) + P(X = 19)

= $\binom{20}{17} \times 0.90^{17} \times (1-0.90)^{20-17}+\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}+\binom{20}{19} \times 0.90^{19} \times (1-0.90)^{20-19}$