Answer: (x-6)(x+1)
Step-by-step explanation:
Sum : x2 − 5 x- 6
x2 + x-=6 x − 6
factor pair: x 2 + x − 6 x − 6
x( x+ 1 ) − 6 ( x + 1 )
rewrite from: x (x + 1 ) − 6 ( x + 1 )
(x-6)(x+1)
Answer:
<em>There is no significant difference in the amount of rain produced when seeding the clouds.</em>
Step-by-step explanation:
Assuming that the amount of rain delivered by thunderheads follows a distribution close to a normal one, we can formulate a hypothesis z-test:
<u>Null Hypothesis
</u>
: Average of the amount of rain delivered by thunderheads without seeding the clouds = 300 acrefeet.
<u>Alternative Hypothesis
</u>
: Average of the amount of rain delivered by thunderheads by seeding the clouds > 300 acrefeet.
This is a right-tailed test.
Our z-statistic is
We now compare this value with the z-critical for a 0.05 significance level. This is a value
such that the area under the Normal curve to the left of
is less than or equal to 0.05
We can find this value with tables, calculators or spreadsheets.
<em>In Excel or OpenOffice Calc use the function
</em>
<em>NORMSINV(0.95)
</em>
an we obtain a value of
= 1.645
Since 1.2845 is not greater than 1.645 we cannot reject the null, so the conclusion that can be drawn when the significance level is 0.05 is that there is no significant difference in the amount of rain produced when seeding the clouds.
Answer:
Correct option: (a) 0.1452
Step-by-step explanation:
The new test designed for detecting TB is being analysed.
Denote the events as follows:
<em>D</em> = a person has the disease
<em>X</em> = the test is positive.
The information provided is:

Compute the probability that a person does not have the disease as follows:

The probability of a person not having the disease is 0.12.
Compute the probability that a randomly selected person is tested negative but does have the disease as follows:
![P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%29%3DP%28X%5E%7Bc%7D%7CD%29P%28D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%20P%28D%29%5C%5C%3D%5B1-0.97%5D%5Ctimes%200.88%5C%5C%3D0.03%5Ctimes%200.88%5C%5C%3D0.0264)
Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:
![P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%5E%7Bc%7D%29%3DP%28X%5E%7Bc%7D%7CD%5E%7Bc%7D%29P%28D%5E%7Bc%7D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%7B1-%20P%28D%29%5D%5C%5C%3D0.99%5Ctimes%200.12%5C%5C%3D0.1188)
Compute the probability that a randomly selected person is tested negative as follows:


Thus, the probability of the test indicating that the person does not have the disease is 0.1452.
Before we get started, it is good to remember PEMDAS - the acronym that tells you the order to carry out equations.
P = parentheses
E = exponents
M = multiplication
D = division
A = addition
S = subtraction
Knowing this, we should start by doing the equations <em>within </em>each parentheses.

So we did the addition and subtraction pieces within each group leaving us with the above equation. Now let's multiply:

20 is the answer.
The last awnser is 15x + 2 and the first one is explained below:
65 X 1.5 = 97.5
97.5 + 95 = 192.5
192.5 is awnser
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