Can someone please help me on this problem I’m struggling

Graph the line with the equation y = 1/6x-5<br><br> PLEASE HELP

quester [9]

Answer:

Step-by-step explanation:

5 0

3 years ago

Find the value of x. <br> pls solve

dmitriy555 [2]

the answer is x = 120 degrees

8 0

4 years ago

Jeanine Baker makes floral arrangements. She has 15 different cut flowers and plans to use 5 of them. How many different selecti

AlladinOne [14]

For the first flower she has a choice of 15. For the second 14 (15 minus the one that was already used). For the third 13, and so on.

Each of the first 15 can be match with any of the second 14. Any of the first pairs can be matched with ant of the third 13 and so on.

This gives us

Number of Selections = Choices for Flower 1 x Choices for Flower 2 x Choices for Flower 3 x Choices for Flower 4 x Choices for Flower 5

$N_{s}=f_{1} \times f_{2} \times f_{3} \times f_{4} \times f_{5}$

$N_{s}=15 \times 14 \times 13 \times 12 \times 11$

$N_{s}=360,360$

You could also calculate this as the permutation $_{15}P_{5}=\frac{15!}{(15-5)!}$

7 0

3 years ago

Read 2 more answers

A financial talk show host claims to have a 55.3 % success rate in his investment recommendations. You collect some data over th

baherus [9]

Answer:

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Step-by-step explanation:

For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

$p = 0.553, n = 10$

If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:

This is

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$