I need help please
1 answer:
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Answer:
Bag of windflower bulbs costs $8.50
Package of crocus bulbs costs $17.60
Step-by-step explanation:
Let $x be the price of one bag of windflower bulbs and $y be the price of one package of crocus bulbs.
1. Mark sold 2 bags of windflower bulbs for $2x and 5 packages of crocus bulbs for $5y. In total he earned $(2x+5y) that is $105. So,
2x+5y=105
2. Julio sold 9 bags of windflower bulbs for $9x and 5 packages of crocus bulbs for $5y. In total he earned $(9x+5y) that is $164.50. So,
9x+5y=164.50
3. You get the system of two equations:
From the first equation
Substitute it into the second equation:
9x+105-2x=164.50
7x=164.50-105
7x=59.5
x=$8.50
So,
5y=105-2·8.5
5y=105-17
5y=88
y=$17.60
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(X
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) =
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
I hope you have a SUPER day!