1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
maria [59]
3 years ago
14

I wanna know what is 42/24 simplified

Mathematics
1 answer:
borishaifa [10]3 years ago
8 0
This is my awnser
7
--=1.75000
4
You might be interested in
Not all visitors to a certain company's website are customers. In fact, the website administrator estimates that about 5% of all
Gnom [1K]

Answer:

0.0135 = 1.35% probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

Step-by-step explanation:

For each visitor of the website, there are only two possible outcomes. Either they are looking for the website, or they are not. The probability of a customer being looking for the website is independent of other customers. This means that the binomial probability distribution is used 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.

5% of all visitors to the website are looking for other websites.

So 100 - 5 = 95% are looking for the website, which means that p = 0.95

Find the probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

This is P(X = 2) when n = 4. So

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

P(X = x) = C_{4,2}.(0.95)^{2}.(0.05)^{2} = 0.0135

0.0135 = 1.35% probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

5 0
2 years ago
which of the following ordered pairs could be placed in the table below and still have the relation qualify as a linear function
mr_godi [17]
(1,31)
Subtract 2 from x, add 8 to y
8 0
3 years ago
8(5-32) equal to and explain <br> please help
Gre4nikov [31]
You would use order of operations: PEMDAS
P(parenthesis) E(exponents) MD(multiplication/division) AS(addition/subtraction)
with MD and AS order doesnt matter.
8(5-32) you would start with inside the Parenthesis for "P" so (5-32)=(-27)
next you would go to the E but because you dont have an exponent you go to the next step with is the "MD" you have multiplication so next would be 8(-27) and 8 multiplied by -27 is: 8(-27)= -216
ANSWER: -216
4 0
3 years ago
Read 2 more answers
Anyone help please....
MissTica

mean= 4.1

median= 1,2,3,4,5,6

mode= 6

Step-by-step explanation:

4 0
3 years ago
ASAP PLS
VikaD [51]
Man in gonna tell you something but it might be wrong. about 1 or 2 i think?? i'm not good at this
3 0
3 years ago
Other questions:
  • How many times can 2 go into 238​
    10·1 answer
  • What is the y-intercept of the line -3 + 4y = 12
    15·1 answer
  • The difference of seventeen and four times a number
    11·1 answer
  • Find the mean. 217, 230, 214, 227, 196, 235, 220, 224, 208, 209, 191, 205, 184, 214, 219, 208, 227, 194, 228, 186, 201, 239
    9·1 answer
  • A dataset from a government survey contains variables for 87 adults age from 18 to 85. The variables include iron level ( µ g/d
    7·1 answer
  • Karen is trying to determine how long her 4-year-old daughter should sit in time-out for deliberately pouring her juice on the f
    8·1 answer
  • 4 thousands equal how many hundreds
    15·2 answers
  • Need help with my school
    6·1 answer
  • Can someone help me with this thanks
    12·2 answers
  • What is Jamal’s weight in kilograms if he weighs 156 pounds? Round to the nearest tenth if necessary. Jamal weighs 34.7 kilogram
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!