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kozerog [31]
3 years ago
15

Some one please help

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
7 0

Answer: d

|4x + 1| ≤ 5

⇒ - 5 ≤ 4x + 1 ≤ 5

⇔ -6 ≤ 4x ≤ 4

⇔ -1.5 ≤ x ≤ 1

⇒ x ∈ [-1.5; 1]

Step-by-step explanation:

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Solve the compound inequality and graph<br> Please help! :)
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Answer:

Step-by-step explanation:

x+4≥10 and x-6>-15

x≥10-4 and x>-15+6

x≥ 6 and x>-9

combining the two

x>-9

8 0
3 years ago
What is theory of demand?<br>​
Yuri [45]

Answer:

Demand theory is an economic principle relating to the relationship between consumer demand for goods and services and their prices in the market. ... As more of a good or service is available, demand drops and so does the equilibrium price.

4 0
3 years ago
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I need to know if this Answer is correct. please tell me the answer. if it's wrong. Thanks!!!
yaroslaw [1]
Here, as I can see, you are adding fractions, and making sure you have a common denominator. 

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13) is correct

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3 years ago
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Medical scientists study the effect of acute infection on tissue-specific immunity. In a collection of experiments under the sam
Serjik [45]

Answer:

The 95% confidence interval  for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

Z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

In a collection of experiments under the same conditions, 44 of 75 mice test positive for lymphadenopathy. This means that n = 75 and \pi = \frac{44}{75} = 0.586.

Compute a 95% confidence interval for the true proportion of mice that will test positive under similar conditions.

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 - 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.5291

The upper limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 + 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.6429

The 95% confidence interval  for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).

7 0
3 years ago
I did not do as good as I hoped on my math model :(
pochemuha
I’m sorry! I hope you do better next time.
8 0
3 years ago
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