How to represent the number 50,679

The zeroes of f(x) = x3 + 3x2 + 2x

Bond [772]

Answer:

the answers are -2,-1,0

Step-by-step explanation:

6 0

3 years ago

Makes 12 cinnamon rolls needs to make 30

nignag [31]

Oh thanks Pls help ASAP i don’t have time!!!!!

8 0

3 years ago

Read 2 more answers

I have 1 litre of water in a jug. On the way to my table I spill some. I have 780ml left. How much have I lost?

irina1246 [14]

Step-by-step explanation:

1 litre = 1000ml

780ml left

lost amount = 1000 - 780 = 220ml

3 0

2 years ago

A guy-wire is attached from the ground to the top of a pole for support. If the angle of elevation to the pole is 67° and the wi

zubka84 [21]

100 and 70 for the 67 and 137

4 0

3 years ago

7.109) The leading brand of dishwasher detergent has a 30% market share. A sample of 25 dishwasher detergent customers was taken

Rina8888 [55]

Answer:

0.9021 = 90.21% probability that 10 or fewer customers choose the leading brand

Step-by-step explanation:

For each customer, there are only two possible outcomes. Either they choose the leading brand, or they do not. The probability of a customer choosing the leading brand is independent of any other customer, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The leading brand of dishwasher detergent has a 30% market share.

This means that $p = 0.3$

A sample of 25 dishwasher detergent customers was taken.

This means that $n = 25$

a. What is the probability that 10 or fewer customers choose the leading brand?

This is:

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.3)^{0}.(0.7)^{25} = 0.0001$

$P(X = 1) = C_{25,1}.(0.3)^{1}.(0.7)^{24} = 0.0014$

$P(X = 2) = C_{25,2}.(0.3)^{2}.(0.7)^{23} = 0.0074$

$P(X = 3) = C_{25,3}.(0.3)^{3}.(0.7)^{22} = 0.0243$

$P(X = 4) = C_{25,4}.(0.3)^{4}.(0.7)^{21} = 0.0572$

$P(X = 5) = C_{25,5}.(0.3)^{5}.(0.7)^{20} = 0.1030$

$P(X = 6) = C_{25,6}.(0.3)^{6}.(0.7)^{19} = 0.1472$

$P(X = 7) = C_{25,7}.(0.3)^{7}.(0.7)^{18} = 0.1712$

$P(X = 8) = C_{25,8}.(0.3)^{8}.(0.7)^{17} = 0.1651$

$P(X = 9) = C_{25,9}.(0.3)^{9}.(0.7)^{16} = 0.1336$

$P(X = 10) = C_{25,10}.(0.3)^{10}.(0.7)^{15} = 0.0916$

Then

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0001 + 0.0014 + 0.0074 + 0.0243 + 0.0572 + 0.1030 + 0.1472 + 0.1712 + 0.1651 + 0.1336 + 0.0916 = 0.9021$

0.9021 = 90.21% probability that 10 or fewer customers choose the leading brand

3 0

2 years ago