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4 years ago

Help Me with theses four questions please ty

Mathematics

2 answers:

Alex_Xolod [135]4 years ago

5 0

1. 135.4896pi m^2
2. 39.69pi cm^2
3.1017 mm^2
4.225pi cm^2

The formula for these problems is

kondor19780726 [428]4 years ago

4 0

1. 135.4896pi m^2
2. 39.69pi cm^2
3.1017 mm^2
4.225pi cm^2

The formula for these problems is $A= \pi r^{2}$

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Between which two numbers does the root square 5 lie

docker41 [41]

Answer:

Between 2 and 3

Step-by-step explanation:

The square root of 5 equals roughly 2.3, so it belongs between 2 and 3 on a number line.

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3 years ago

Find the value of i182. A) 1 B) i C) ?1 D) ?i

elena-14-01-66 [18.8K]

What are the question marks?
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Simplify the expression using the definition of an imaginary number.

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3 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

Write the slope-intercept form of the equation of the line that has the same y-intercept as the black line shown and passes thro

Mazyrski [523]

Answer:

y = 0.5x + 2

Step-by-step explanation:

the y-intercept is 2, as shown in the graph

the slope is 0.5. this is because while the x-value goes up by 2, the y-value goes up by one (from 0, 2 to 2, 3).

using the slope formula of rise / run, you get 1/2, which is equal to 0.5

5 0

3 years ago

When writing 3.7272727272... as a ratio of integers, what is the common ratio?

Law Incorporation [45]

The answer is: "41 / 11 " ; or, write as: " 41 : 11 " .
__________________________________________________________
Explanation:
__________________________________________________________
72/99 = (72÷9) / (99÷9) = 8/ 11 .

So, 3.727272727272... = "3 $\frac{8}{11}$ " .

→ "3 $\frac{8}{11}$ " = [(11 *3) + 8] / 11 = (33 + 8) / 11 = 41 / 11 .
__________________________________________________________
"41 / 11" cannot be simplified any further; so the answer is:
__________________________________________________________
"41 / 11 " ; or, write as: " 41 : 11 " .
__________________________________________________________

5 0

3 years ago