[tex]solution \\ {2x}^{2} - x + ( - x - {2x}^{2} - 2) \\ = {2x}^{2} - x - x - {2x}^{2} - 2 \\ = {2x}^{2} - {2x}^{2} - x - x - 2 \\ = - 2x - 2[/tex]
Hope it helps
Good luck on your assignment
Answer:
[tex] - 2x - 2[/tex]
Step-by-step explanation:
[tex]2 {x}^{2} - x + ( - x - 2 {x}^{2} - 2) \\ 2 {x}^{2} - x - x - 2 {x}^{2} - 2 \\ 2 {x}^{2} - 2 {x}^{2} - x - x - 2 \\ - 2x - 2[/tex]
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Keith Rollag (2007) noticed that coworkers evaluate and treat "new" employees differently from other staff members. He was interested in how long a new employee is considered "new" in an organization. He surveyed four organizations ranging in size from 34 to 89 employees. He found that the "new" employee status was mostly reserved for the 30% of employees in the organization with the lowest tenure.
A) In this study, what was the real range of employees hired by each organization surveyed?
B) What was the cumulative percent of "new" employees with the lowest tenure?
Answer:
a) Real range of employees hired by each organization surveyed = 56
b) The cumulative percent of "new" employees with the lowest tenure = 30%
Step-by-step explanation:
a) Note: To get the real range of employees hired by each organization, you would do a head count from 34 to 89 employees. This means that this can be done mathematically by finding the difference between 34 and 89 and add the 1 to ensure that "34" is included.
Real range of employees hired by each organization surveyed = (89 - 34) + 1
Real range of employees hired by each organization surveyed = 56
b) It is clearly stated in the question that the "new" employee status was mostly reserved for the 30% of employees in the organization with the lowest tenure.
Therefore, the cumulative percent of "new" employees with the lowest tenure = 30%
solve for x
2x/3 + 2 = 16
Answer:
2x/3 + 2= 16
=21
Step-by-step explanation:
Standard form:
2
3
x − 14 = 0
Factorization:
2
3 (x − 21) = 0
Solutions:
x = 42
2
= 21
In a certain community, eight percent of all adults over age 50 have diabetes. If a health service in this community correctly diagnosis 95% of all persons with diabetes as having the disease and incorrectly diagnoses ten percent of all persons without diabetes as having the disease, find the probabilities that:
Complete question is;
In a certain community, 8% of all people above 50 years of age have diabetes. A health service in this community correctly diagnoses 95% of all person with diabetes as having the disease, and incorrectly diagnoses 10% of all person without diabetes as having the disease. Find the probability that a person randomly selected from among all people of age above 50 and diagnosed by the health service as having diabetes actually has the disease.
Answer:
P(has diabetes | positive) = 0.442
Step-by-step explanation:
Probability of having diabetes and being positive is;
P(positive & has diabetes) = P(has diabetes) × P(positive | has diabetes)
We are told 8% or 0.08 have diabetes and there's a correct diagnosis of 95% of all the persons with diabetes having the disease.
Thus;
P(positive & has diabetes) = 0.08 × 0.95 = 0.076
P(negative & has diabetes) = P(has diabetes) × (1 –P(positive | has diabetes)) = 0.08 × (1 - 0.95)
P(negative & has diabetes) = 0.004
P(positive & no diabetes) = P(no diabetes) × P(positive | no diabetes)
We are told that there is an incorrect diagnoses of 10% of all persons without diabetes as having the disease
Thus;
P(positive & no diabetes) = 0.92 × 0.1 = 0.092
P(negative &no diabetes) =P(no diabetes) × (1 –P(positive | no diabetes)) = 0.92 × (1 - 0.1)
P(negative &no diabetes) = 0.828
Probability that a person selected having diabetes actually has the disease is;
P(has diabetes | positive) =P(positive & has diabetes) / P(positive)
P(positive) = 0.08 + P(positive & no diabetes)
P(positive) = 0.08 + 0.092 = 0.172
P(has diabetes | positive) = 0.076/0.172 = 0.442
Using formula:
[tex]P(\text{diabetes diagnosis})\\[/tex]:
[tex]=\text{P(having diabetes and have been diagnosed with it)}\\ + \text{P(not have diabetes and yet be diagnosed with diabetes)}[/tex]
[tex]=0.08 \times 0.95+(1-0.08) \times 0.10 \\\\=0.08 \times 0.95+0.92 \times 0.10 \\\\=0.076+0.092\\\\=0.168[/tex]
[tex]\text{P(have been diagnosed with diabetes)}[/tex]:
[tex]=\frac{\text{P(have diabetic and been diagnosed as having insulin)}}{\text{P(diabetes diagnosis)}}[/tex]
[tex]=\frac{0.08\times 0.95}{0.168} \\\\=\frac{0.076}{0.168} \\\\=0.452\\[/tex]
Learn more about the probability:
brainly.com/question/18849788
Please answer this correctly
Answer:
101-120=4
Step-by-step explanation:
All that you need to do is count how many data points fall into this category. In this case, there are four data points that fall into the category of 101-120 pushups
111111105113Therefore, the answer to the blank is 4. If possible, please mark brainliest.
Answer:
There are 4 numbers between 101 and 120.
Step-by-step explanation:
101-120: 105, 111, 111, 113 (4 numbers)