All categories

3 years ago

I just need help with number 36 pleasee

Mathematics

1 answer:

Molodets [167]3 years ago

3 0

i dont know sorry lkdkjdndn huh;iuh iuh hi uhi uh h h ih iu uh iuhudffh;uf h

You might be interested in

What is the answer to the following question?

velikii [3]

Choice B the cos is the adjacent side of the angle over the hypotenuse

8 0

4 years ago

Approximately how many centimeters are in 54 ft?

dedylja [7]

Answer:

d) 54 ft = 1645.92 cm

Step-by-step explanation:

Given : 54 ft.

To find : Approximately how many centimeters are in 54 ft.

Solution : We have given 54 ft.

We know

1 ft = 30.48 cm .

54 ft = 54 * 30.48 cm.

54 ft = 1645.92 cm.

Therefore, d) 54 ft = 1645.92 cm.

5 0

3 years ago

A die is tossed. find the odds against rolling a number greater than 2.

alex41 [277]

Umm what do you mean can you explain more of it 2 me plz<span />

3 0

4 years ago

Can I get help with 1 & 2??

Rashid [163]

A. g(10)= -29
b.f(3)=16
c.h(-2)=-6
d.j(7)=23
e.h(a)=12/a
f. g(b+c)=-3b-3c+1 ...etc I'm bored

3 0

4 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

3 years ago