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
katrin [286]
3 years ago
15

If x

%20x%20%20%2B%20%20%5Cfrac%7B1%7D%7B%3Fx%7D%20%20%3D%20" id="TexFormula1" title=" x = \sqrt[?]{5 } + \sqrt{3} + x + \frac{1}{?x} = " alt=" x = \sqrt[?]{5 } + \sqrt{3} + x + \frac{1}{?x} = " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
Andru [333]3 years ago
4 0

Answer:

10 − 3 x = 2 x + 5 \sqrt{10-3x}=\sqrt{2x+5} 10−3x ​=2x+5 ​square root of, 10, minus, 3, x, end square root, equals, square root of, 2, x, plus, 5, end square root.

Step-by-step explanation:

sasho [114]3 years ago
3 0

Answer:

bh

Step-by-step explanation:

34

You might be interested in
Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in
arsen [322]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in the summer of 2014, Google's Chrome browser exceeded a 20% market share for the first time, with a 20.37% share of the browser market (Forbes website, December ). For a randomly selected group of 20 Internet browser users, answer the following questions.

a. Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser (to 4 decimals). For this question, if you compute the probability manually, make sure to carry at least six decimal digits in your calculations.

b. Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser (to 4 decimals).

c. For the sample of Internet browser users, compute the expected number of Chrome users (to 3 decimals).

d. For the sample of Internet browser users, compute the variance and standard deviation for the number of Chrome users (to 3 decimals).

Answer:

a. P(x = 8) = 0.024273

b. P(x ≥ 3) = 0.8050

c. E(x) = 4.074

d. var(x) = 3.244 and SD = 1.801

Step-by-step explanation:

The given problem can be solved using the binomial distribution

P(x) = ⁿCₓ pˣ (1 - p)ⁿ⁻ˣ  

Where n is the number of trials, x is the variable of interest and p is the probability of success.  

For the given problem,

Probability of success = p = 20.37% = 0.2037

Probability of failure = q = 1 - p = 1- 0.2037 = 0.7963

Number of trials  = n = 20

a. Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

In this case, we have x = 8

P(x = 8) = ²⁰C₈×0.2037⁸×(1 - 0.2037)²⁰⁻⁸  

P(x = 8) = 125970×0.2037⁸×0.7963¹²

P(x = 8) = 0.024273

b. Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser (to 4 decimals).

P(x ≥ 3) = 1 - P(x < 2)

But we know that

P(x < 2) = P(x = 0) + P(x = 1) + P(x = 2)

So,

P(x ≥ 3) = 1 - [ P(x = 0) + P(x = 1) + P(x = 2) ]

For P(x = 0):

Here we have x = 0, n = 20 and p = 0.2037

P(x = 0) = ²⁰C₀×0.2037⁰×(1 - 0.2037)²⁰⁻⁰

P(x = 0) = 0.0105

For P(x = 1):

Here we have x = 1, n = 20 and p = 0.2037

P(x = 1) = ²⁰C₁×0.2037¹×(1 - 0.2037)²⁰⁻¹

P(x = 1) = 0.0538

For P(x = 2):

Here we have x = 2, n = 20 and p = 0.2037

P(x = 2) = ²⁰C₂×0.2037²×(1 - 0.2037)²⁰⁻²

P(x = 2) = 0.1307

Finally,

P(x ≥ 3) = 1 - [P(x = 0) + P(x = 1) + P(x = 2)]  

P(x ≥ 3) = 1 - [ 0.0105 + 0.0538  + 0.1307]  

P(x ≥ 3) = 1 - [0.1950]

P(x ≥ 3) = 0.8050

c. For the sample of Internet browser users, compute the expected number of Chrome users (to 3 decimals).

The expected number of Chrome users is given by

E(x) = n×p

Where n is the number of trials and p is the probability of success

E(x) = 20×0.2037

E(x) = 4.074

d. For the sample of Internet browser users, compute the variance and standard deviation for the number of Chrome users (to 3 decimals).

The variance for the number of Chrome users is given by

var(x) = n×p×q

Where n is the number of trials and p is the probability of success and q is the probability of failure.

var(x) = 20×0.2037×0.7963

var(x) = 3.244

The standard deviation for the number of Chrome users is given by

SD = √(n×p×q)

SD = √var(x)

SD = √(3.244)

SD = 1.801

5 0
3 years ago
Hailey planted
IrinaVladis [17]

Answer:

18 seeds

Step-by-step explanation:

3 x 6 = 18 seeds

3/5 x 30 = 18 seeds

Hope that helps!

7 0
2 years ago
If you answer this I will give you brainiest!
Helga [31]

Answer:

1 54 2 132 3 334.58 4 103.18

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Find the value of x and y.
Luden [163]

Answer:

I believe it is A......

4 0
3 years ago
Read 2 more answers
The board of directors of a company knows that the probability that carbon emissions from the company’s factory exceed the permi
jekas [21]

the real answer is D.  0.9132

3 0
3 years ago
Read 2 more answers
Other questions:
  • Math second grade (plz help me)
    10·2 answers
  • Stella is doing her math worksheets. It takes her 1 hour 20 minutes to complete one worksheet with 12 questions.
    7·1 answer
  • Which of the following pairs of expressions could represent consecutive odd numbers?
    8·2 answers
  • HELP ME YOULL GET 25 POINTS!!!!!!!!
    5·1 answer
  • D= 5/6 (F+G) solve for G
    11·1 answer
  • Examine the steps shown for testing whether or not x=1 is a zero for the function f(x)=x^3-3x+2. Identify the first line in whic
    9·1 answer
  • What is (2x)^3<br> I need help
    14·2 answers
  • Find the slope of the line passing through the points<br> (-7,-7) and (-3,6)
    8·1 answer
  • Yo can i get some help no links or pictures only say the truth or get reported
    11·2 answers
  • For every 2 hours, approximately $1,302 worth of product is sold to consumers. How much money will the company make from selling
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!