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RideAnS [48]
3 years ago
15

Presley is organizing the school book sale. Each stall will have 40 books on the shelf and 2 boxes of books with b books inside.

Presley also has 200 books in inventory to refill the shelves as books sell. This situation can be written as the following expression:
40s + (2b)s + 200

How many books should Presley have ready if she plans to open 8 stalls and have 12 books in each box?
A) 520 B) 712 C) 820 D) 1,622
Mathematics
2 answers:
kow [346]3 years ago
6 0

B. 712 is the answer


lukranit [14]3 years ago
5 0
40(8) + (2(12))(8) + 200
320 + 24(8) + 200
520 + 192
712
Answer: B. 712
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A researcher plants 22 seedlings. After one month, independent of the other seedlings, each seedling has a probability of 0.08 o
Andrews [41]

Answer:

E(X₁)= 1.76

E(X₂)= 4.18

E(X₃)= 9.24

E(X₄)= 6.82

a. P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= 0.00022

b. P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= 0.000001

c. P(X₁≤2) = 0.7442

Step-by-step explanation:

Hello!

So that you can easily resolve this problem first determine your experiment and it's variables. In this case, you have 22 seedlings (n) planted and observe what happens with the after one month, each seedling independent of the others and has each leads to success for exactly one of four categories with a fixed success probability per category. This is a multinomial experiment so I'll separate them in 4 different variables with the corresponding probability of success for each one of them:

X₁: "The seedling is dead" p₁: 0.08

X₂: "The seedling exhibits slow growth" p₂: 0.19

X₃: "The seedling exhibits medium growth" p₃: 0.42

X₄: "The seedling exhibits strong growth" p₄:0.31

To calculate the expected number for each category (k) you need to use the formula:

E(XE(X_{k}) = n_{k} * p_{k}

So

E(X₁)= n*p₁ = 22*0.08 = 1.76

E(X₂)= n*p₂ = 22*0.19 = 4.18

E(X₃)= n*p₃ = 22*0.42 = 9.24

E(X₄)= n*p₄ = 22*0.31 = 6.82

Next, to calculate each probability you just use the corresponding probability of success of each category:

Formula: P(X₁, X₂,..., Xk) = \frac{n!}{X_{1}!X_{2}!...X_{k}!} * p_{1}^{X_{1}} * p_{2}^{X_{2}} *.....*p_{k}^{X_{k}}

a.

P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= \frac{22!}{3!4!6!} * 0.08^{3} * 0.19^{4} * 0.42^{6}\\ = 0.00022

b.

P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= \frac{22!}{5!5!7!} * 0.08^{5} * 0.19^{5} * 0.31^{7}\\ = 0.000001

c.

P(X₁≤2) = \frac{22!}{0!} * 0.08^{0} * (0.92)^{22} + \frac{22!}{1!} * 0.08^{1} * (0.92)^{21} + \frac{22!}{2!} * 0.08^{2} * (0.92)^{20} = 0.7442

I hope you have a SUPER day!

8 0
2 years ago
Please Help!!!!!!! 10 Points!!!!!!
timurjin [86]

Answer:

tyyegwgrg

Step-by-step explanation:

htfrbgregbrfehetbfdb

4 0
2 years ago
PLEASE HELP WILL MARK BRAINLIEST!!!!!
trapecia [35]

The numbers inside the parenthesis is the value of x, uses the equation associated with the value of x and solve.

H(1) Since 1 fits 1≤x≤3 use x^3

x^3 1^3 = 1

h(1) = 1

h(4) 4 fits x >3, so use 5

h(4) = 5

5 0
2 years ago
On a coordinate plane, point AS
nlexa [21]

Answer:

√1370 or about 37.01

Step-by-step explanation:

use distance formula

√(x2-x1)^2+(y2-y1)^2

sqrt(8-7)^2+(41-4)^2

sqrt(1)^2+(37)^2

sqrt 1+1369

sqrt 1370

= about 37. 01

3 0
2 years ago
Find all values of x in the interval [0,2π] that satisfy the equation sin2x=cosx. enter your answers as a list of numbers separa
yuradex [85]

\sin(2x)  =  \cos(x)  \\ 2 \sin(x)  \cos(x) =  \cos(x)  \\  \cos(x) (2 \sin(x)  - 1) = 0
so
\cos(x)  = 0 >  >  > x =  \frac{\pi}{2}  + k\pi \\  \sin(x)  =  \frac{1}{2}  >  >  > x =  \frac{\pi}{6}  + 2k\pi \\ or \: x =  \frac{5\pi}{6}  + 2k\pi
x =  \frac{\pi}{2} or \frac{3\pi}{2} \\  x =  \frac{\pi}{6} or \frac{5\pi}{6}
6 0
3 years ago
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