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Tamiku [17]
3 years ago
12

Let X be a normal random variable with mean 40 and standard deviation 2. Find P(X < 28).

Mathematics
1 answer:
Vikki [24]3 years ago
5 0

Answer:

0

Step-by-step explanation:

Given :

Mean = 40 ; Standard deviation = 2

P(x < 28)

Obtain the Zscore :

Z = (x - mean) / standard deviation

P(Z < (28 - 40) / 2))

Z = (28 - 40) / 2

Z = - 12 / 2

Z = - 6

P(Z < - 6) = 0.000000

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Tickets to a movie cost $7.25 for adults and $5.50 for students a group of friends purchased 8 tickets for $52.75 how many adult
irina1246 [14]

Answer:

The number of adults tickets and student tickets purchased is 5 and 3.

Given that,

Tickets to a movie cost $7.25 for adults and $5.50 for students.

A group of friends purchased 8 tickets for $52.75.

Here we assume the no of adult tickets and no of student tickets be x and y.

Based on the above information, the calculation is as follows:

7.25x + 5.50y = 52.75 .............(i)

x + y = 8

x = 8 - y..................(2)

Now put the x value in the equation (1)

So,  

7.25(8-y) + 5.50y = 52.75

7.25 × 8 - 7.25y + 5.50y = 52.75

58 - 1.75y = 52.75

5.25 = 1.75y

y = 3

So,  

x = 8 - 3

= 5

Therefore we can conclude that the number of adults tickets and student tickets purchased is 5 and 3.

-Hope this helps<3

4 0
2 years ago
For which inequality is {x | x R, x &gt; 6} the solution set?
CaHeK987 [17]

Answer:

C

Step-by-step explanation:

<em>B: 2X + 12 > 3</em>

2x > -9

x> -9/2

<em>C: 2X - 5 > 7</em>

2x > 12

x > 6

<em>D: 2x - 6 > 24</em>

2x > 30

x > 15

Hope this helps ^-^

6 0
3 years ago
A bacteria colony increases in size at a rate of 4.0581e1.6t bacteria per hour. If the initial population is 36 bacteria, find t
Harrizon [31]

Answer:

1560

Step-by-step explanation:

The rate of Increase of the population (P) of the bacteria is given as:

\frac{dP}{dt} =4.058e^{1.6t}

dP =4.058e^{1.6t}}dt\\Taking\: Integrals\\\int dP =4.058 \int e^{1.6t}dt\\P(t)=\frac{4.058}{1.6} (e^{1.6t} +K)\\P(t)=2.53625 (e^{1.6t} +K)

Where k is a constant of Integration.

At t=0, P(t)=36

36=2.53625 (e^{1.6*0} +K)\\36=2.53625 (e^{0} +K)\\36=2.53625 (1 +K)\\36=2.53625 +2.53625K\\K=13.19

Therefore:

P(t)=2.53625 (e^{1.6t} +13.19)\\At \: t=4\\P(4)=2.53625 (e^{1.6*4} +13.19)\\=1559.88\\P(4)=1560

3 0
3 years ago
Find all zeros<br>f(x)=x^4+6x^3+173x^2+332x+164​
zheka24 [161]
The zeros are (-1,0)
6 0
3 years ago
Short on time, please help!
kvv77 [185]
56.3% hope it helped ok
3 0
3 years ago
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