All categories

3 years ago

Which of the following numbers is less than 2/3? F. -1/2 G. 1/3 H. -7/8 J. -1/6

Mathematics

2 answers:

loris [4]3 years ago

5 0

All of them. F, H and J are negative numbers, and as such they are less than any positive number.

Finally, 1/3 is trivially less than 2/3.

stiks02 [169]3 years ago

4 0

Answer:

All of them are.

Step-by-step explanation:

F. H. J. are all negative and G. is just smaller than 2/3

You might be interested in

AB passes through A(-3, 0) and B(-6, 5). What is the equation of the line that passes through the origin and is parallel to AB?

dalvyx [7]

Answer:

i think the correct answer is E.

5 0

3 years ago

Read 2 more answers

Find en equation of the tangent line to the curve

valkas [14]

Y=MX+C C=1 0=m(1)+1 m=-1 Y=-X+1

7 0

3 years ago

your basketball team made 10 three pointers 8 two pointers and 12 one point foul shots your openents made 8 three pointers 12 tw

inn [45]

Answer:

The opponents won by 1 points

Step-by-step explanation:

Your team (10*3)+(8*2)+(12*1)=58

Opponents (8*3)+(12*2)+(11*1)=59

8 0

3 years ago

A golf course charges $16 for a package including the full 18-hole course. The course also sells buckets of

MA_775_DIABLO [31]

16x + 21y = 555

Step-by-step explanation:

Let x be the no. of 18-hole course

And y be the no. of golf balls

16x + 21y = 555

6 0

3 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