Let’s say it takes time t hours for the interception.
In that time the carrier travels 28t nm west and the helicopter 130t nm.
Now we use the sine rule to find the angle x which lies between the distance 175 and 130t.
sin(x)/28t=sin35/130t. So sin(x)=28sin35/130=14sin35/65=0.1235 and x=7.0964 degrees.
Therefore the helicopter should use a bearing of 35+7.0964=42.0964 degrees north of east or 47.9 degrees east of north approx.

8 0

3 years ago

Please help!!!!.................,

dangina [55]

Answer:

I think it's 45 lmk if I'm wrong lol

5 0

3 years ago

12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests

Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using 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:

12% of children are nearsighted. This means that $p = 0.12$ .

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so $n = 18$