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Anna [14]
3 years ago
14

Find 4+(−1 2/3). Write your answer as a fraction in simplest form.

Mathematics
2 answers:
inysia [295]3 years ago
8 0
4/1-12/3
Find LCM
LCM=3
12-12/3
=0/3
=0
Naddik [55]3 years ago
6 0

Answer:2 1/3

make four an fraction and we get 12/3- 1 2/3

ans 2 1/3

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Solve: e^2x + 5 = 4<br> help pls
kaheart [24]

Answer:

The solution of the equation is x=\frac{(ln4)-5}{2} ⇒ 3rd answer

Step-by-step explanation:

* Lets explain how to solve this problem

- The function f(x) = e^x is called the (natural) exponential function

- The natural logarithm (㏑), or logarithm to base e, is the inverse

 function to the natural exponential function

∵ e^{2x+5}=4 is an exponential function

∴ We can solve it by using the inverse of e (㏑)

- Remember:

# ln(e)=1

# ln(e^{m})=m(ln(e))=m

- Insert ln in both sides

∴ ln(e^{2x+5})=ln(4)

∵ ln(e^{2x+5})=(2x+5)ln(e)=2x+5

∴ 2x + 5 = ㏑(4)

- Subtract 5 from both sides

∴ 2x = ㏑(4) - 5

- Divide both sides by 2 to find x

∴ x=\frac{ln(4)-5}{2}

* The solution of the equation is x=\frac{(ln4)-5}{2}

4 0
4 years ago
Read 2 more answers
1. Suppose an online retailer has determined that if a potential customer spends more than 5 minutes on their site, that the pro
vredina [299]

Using the binomial distribution, it is found that there is a 0.056 = 5.6% probability that more than 7 will make a purchase.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • There are 12 customers, hence n = 12.
  • The probability of any of them making a purchase is of p = 0.4.

The probability that more than 7 will make a purchase is given by:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12).

Hence:

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

P(X = 8) = C_{12,8}.(0.4)^{8}.(0.6)^{4} = 0.042

P(X = 9) = C_{12,9}.(0.4)^{9}.(0.6)^{3} = 0.012

P(X = 10) = C_{12,10}.(0.4)^{10}.(0.6)^{2} = 0.02

P(X = 11) = C_{12,11}.(0.4)^{11}.(0.6)^{1} \approx 0

P(X = 12) = C_{12,12}.(0.4)^{12}.(0.6)^{0} \approx 0

Then:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.042 + 0.012 + 0.02 + 0 + 0 = 0.056.

0.056 = 5.6% probability that more than 7 will make a purchase.

More can be learned about the binomial distribution at brainly.com/question/24863377

3 0
2 years ago
In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates
Anettt [7]

Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

S = ut + \frac{1}{2} at^{2}

Where S is distance

u is the initial velocity

a is acceleration

and t is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

S = S₁

u = 0 m/s ( The player starts at rest)

a = 2.2 m/s²

Then,

S₁ = 0(t) + \frac{1}{2} (2.2)t^{2}

S₁ = 1.1t^{2}

t^{2} = \frac{S_{1} }{1.1}

For player 2

S = S₂

u = 0 m/s ( The player starts at rest)

a = 1.7 m/s²

Then,

S₂ = 0(t) + \frac{1}{2} (1.7)t^{2}

S₂ = 0.85t^{2}

t^{2} = \frac{S_{2} }{0.85}

Since the time spent by both players is equal, We can  write that

t^{2} = t^{2}

Then,

\frac{S_{1} }{1.1} =  \frac{S_{2} }{0.85}

0.85S_{1} = 1.1S_{2}

From, S₁ + S₂ = 41 m

S₂ = 41  - S₁

Then,

0.85S_{1} = 1.1 ( 41 -S_{1} )

0.85S_{1} = 45.1 - 1.1S_{1}

0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1}  = 23.13 m

This is the distance covered by the first player. We can then put this value into t^{2} = \frac{S_{1} }{1.1} to determine how much time elapses before the players collide.

t^{2} = \frac{S_{1} }{1.1}

t^{2} = \frac{23.13 }{1.1}

t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs

Hence, the time that elapses before the players collide is 4.59 secs

3 0
4 years ago
Help pls yall i will give points
katen-ka-za [31]

Answer:

im gonna tell you its none promise

8 0
3 years ago
Is ((1,4)(3,2)(8,9)(7,6)(3,4) a function
IRINA_888 [86]

Answer:

yes

Step-by-step explanation:

y's can have the same x

5 0
4 years ago
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