How can you find a function that has these roots in a cubic polynomial function?jj

1 answer:

6 0

Step-by-step explanation:by substituting 1 for each of x in the equation. and apply it for problems, if the solution is 0 that means it the solution of x=1

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The manager of a supermarket wants to obtain information about the proportion of customers who dislike a new policy on cashing c

alex41 [277]

Answer:

n = 61 costumers

Step-by-step explanation:

For calculating the number of costumers he should sample we use the next equation:

$n = \frac{z_{1-\alpha/2}^{2}p(1-p)}{E^{2} }$

Where E is the error that we are prepared to accept, in this case E = 0.15

How we don't know the value of p, we can estimate it like p = 0.5

∝ = 1-0.98 = 0.02

1-∝/2 = 0.99

$z_{0.99} = 2.33$

$n = \frac{(2.33^{2})(0.5)(1-0.5)}{0.15^{2} }$

n = 60.32 costumers

n ≈ 61 costumers

4 0

4 years ago

Create a scenario that leads to an inequality of the form ax + b > c.

Semmy [17]

A scenario that might fit the inequality <span>ax + b > c is that a vendor has a quota of sales of c in $ in which when the time she makes above the quota at the end of a period of time, she will receive a bonus. In this case, she has to sell x items worth $a per x sold and the bonus amount would be b$. So in all, we have to balance the units from left and right, which are both in $. A greater inequality is used because the greater the sales or above the quota, the better sign it is indicating good performance in the market.</span>

5 0

3 years ago

Read 2 more answers

This is what I need help on....

OleMash [197]

Problem 1, part A

X = number of times Roberto throws the baseball in the strike zone
X can take on the following values: 0, 1, 2, 3

p = 0.72 = probability of getting the ball in the strike zone (for any given independent trial)
n = 3 = sample size = number of times the baseball is thrown

Binomial Probabilities:

P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 0) = (3 C 0)*(0.72)^(0)*(1-0.72)^(3-0)
P(X = 0) = (3 C 0)*(0.72)^(0)*(0.28)^(3)
P(X = 0) = (1)*(0.72)^(0)*(0.28)^3
P(X = 0) = (1)*(1)*(0.021952)
P(X = 0) = 0.021952
P(X = 0) = 0.022 <--- this value is added to the table (next to k = 0)

P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 1) = (3 C 1)*(0.72)^(1)*(1-0.72)^(3-1)
P(X = 1) = (3 C 1)*(0.72)^(1)*(0.28)^(2)
P(X = 1) = (3)*(0.72)^(1)*(0.28)^2
P(X = 1) = (3)*(0.72)*(0.0784)
P(X = 1) = 0.169344
P(X = 1) = 0.169 <--- this value is added to the table (next to k = 1)

P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 2) = (3 C 2)*(0.72)^(2)*(1-0.72)^(3-2)
P(X = 2) = (3 C 2)*(0.72)^(2)*(0.28)^(1)
P(X = 2) = (3)*(0.72)^(2)*(0.28)^1
P(X = 2) = (3)*(0.5184)*(0.28)
P(X = 2) = 0.435456
P(X = 2) = 0.435 <--- this value is added to the table (next to k = 2)

P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 3) = (3 C 3)*(0.72)^(3)*(1-0.72)^(3-3)
P(X = 3) = (3 C 3)*(0.72)^(3)*(0.28)^(0)
P(X = 3) = (1)*(0.72)^(3)*(0.28)^0
P(X = 3) = (1)*(0.373248)*(1)
P(X = 3) = 0.373248
P(X = 3) = 0.373 <--- this value is added to the table (next to k = 3)

The table will look like what you see in the attached image

-----------------------------
Problem 1, part B

Refer to the table made in part A above. Add up the values in the second column, the P(X = k) column, that correspond to k values of 1 or larger. So basically everything but the first item which corresponds to k = 0

0.169+0.435+0.373 = 0.977

So the probability of at least one of the baseballs hits the strike zone is 0.977

=========================================
Problem 2

Convert each raw x score to a z score

Company A
z = (x-mu)/sigma
z = (260-276)/5.8
z = -2.759
----------------
Company B
z = (x-mu)/sigma
z = (260-252)/3.4
z = 2.353
----------------
The z scores are -2.759 and 2.353 for company A and company B respectively. The value -2.759 is further away from zero compared to 2.353, so company A has a lower chance of producing 260 nails. This is because company B has x = 260 closer to the mean (than compared to company A