4.3 ft 3.7 ft 6.5 ft 39.96 ft2 26 it? 19.98 ft? 12.03 ft?

Can you please answer this for me

Vsevolod [243]

Answer:

you are on the right track

Step-by-step explanation:

the second ones right so you are right :)

3 0

3 years ago

PLEASE HELP!!!! Geometry is so hard and this is due tonight :(

omeli [17]

Cos(68)=z/24
Z=24cos(68)
Find the answer by the calculator

Second question,
Cos(x)=16/34
X=cos^-1(16/34)
Use calculator

7 0

2 years ago

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

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 combinations 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.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

$P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401$

40.1% probability that he will miss at least one of them

7 0

2 years ago

If two objects travel through space along two different curves, it’s often important to know whether they will collide. (Will a

galina1969 [7]

Answer:

Both objects collide at t=3 in the point <9,9,9>

Step-by-step explanation:

Collision Of Moving Objects

Two objects can describe different trajectories in the space. Those trajectories can intersect in one or more points but it doesn't mean they collide. Collision occurs if they are in the same position at the same time. If we know the positions as a function of time of each object, we could try so find if, for a given time, they are in the same position.

The positions of two object are given as

$r1(t)=$

$r2(t)=$

Let's find out if there is at least one value of t that makes both positions to be the same. We can try by equating one of the three coordinates and testing if the value of t make both have the same x,y,z coordinate. Let's try equating the x-components of both

$t^2=4t-3$