Someone plz help me fast for my exam

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(X $E(X_{k}) = n_{k} * p_{k}$

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

3 years ago

What is the largest possible integral value in the domain of the real-valued function

kotegsom [21]

Answer:

Max Value: x = 400

General Formulas and Concepts:

Algebra I

Domain is the set of x-values that can be inputted into function f(x)

Calculus

Antiderivatives
Integral Property: $\int {cf(x)} \, dx = c\int {f(x)} \, dx$
Integration Method: U-Substitution
[Integration] Reverse Power Rule: $\int {x^n} \, dx = \frac{x^{n+1}}{n+1} + C$

Step-by-step explanation:

Step 1: Define

$f(x) = \frac{1}{\sqrt{800-2x} }$

Step 2: Identify Variables

Using U-Substitution, we set variables in order to integrate.

$u = 800-2x\\du = -2dx$

Step 3: Integrate

Define: $\int {f(x)} \, dx$
Substitute: $\int {\frac{1}{\sqrt{800-2x} } } \, dx$
[Integral] Int Property: $-\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx$
[Integral] U-Sub: $-\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du$
[Integral] Rewrite: $-\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du$
[Integral - Evaluate] Reverse Power Rule: $-\frac{1}{2}(2\sqrt{u}) + C$
Simplify: $-\sqrt{u} + C$
Back-Substitute: $-\sqrt{800-2x} + C$
Factor: $-\sqrt{-2(x - 400)} + C$

Step 4: Identify Domain

We know from a real number line that we cannot have imaginary numbers. Therefore, we cannot have any negatives under the square root.

Our domain for our integrated function would then have to be (-∞, 400]. Anything past 400 would give us an imaginary number.

7 0

3 years ago

Simplity |10| Right the answer here please

Scrat [10]

Answer:

$10$

Step-by-step explanation:

The lines surrounding 10 are called positive brackets, meaning the no matter what number is in those brackets, it will always come out of positive.

There is nothing to do with just 10 because we don't need to change anything.

|10| simplified is 10

5 0

3 years ago

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 60 o

krok68 [10]

Answer:

a) 99.7% of the widget weights lie between 45 and 75 ounces

b) 97.2% of the widget weights lie between 50 and 75 ounces

c) 84% of the widget weights lie above 55

Step-by-step explanation:

The Empirical Rule states that:

50% of the values of a measure is above the mean, and the other 50% is below the mean.

99.7% of the values of a measure lie between 3 standard deviations of the mean.

95% of the values of a measure lie between 2 standard deviations of the mean.

68% of the values of a measure lie between 1 standard deviations of the mean.

In this problem, we have that: The widget weights have a mean of 60 ounces and a standard deviation of 5 ounces.

(a) 99.7% of the widget weights lie between

3 standard deviations of the mean, so:

60 - 3*5 = 60 + 3*5 = 45 and 75 ounces

(b) What percentage of the widget weights lie between 50 and 75 ounces?

We have to find the percentage that are below 75 and subtract by the percentage that are below 50. So

75 is 3 standard deviations above the mean. So 99.7% of the measures are below 75.

50 is 2 standard deviations below the mean. So only 5% of the measures that are below the mean are below 50.

So

99.7% - (50%)5% = 99.7% - 2.5% = 97.2%

(c) What percentage of the widget weights lie above 55?

55 is one standard deviation below the mean.

50% of the widget weights are above the mean.

Of the 50% that is below, 68% lie between one standard deviation(So from 55 to 60)

So

50% + 68%(50%) = 84%

3 0

3 years ago

Tony spent $11.24 and bought 5 packs of notebooks and a calculator. How much does a pack of notebooks cost if a calculator costs

Dominik [7]

Each pack of notebooks would be $0.75.

5 0

3 years ago