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sveticcg [70]
3 years ago
12

Please i need help solve for x

Mathematics
2 answers:
____ [38]3 years ago
7 0

Answer:

x <= -5

Step-by-step explanation:

4x + 12 <= -8

4x <= -20

x <= -5

malfutka [58]3 years ago
3 0

Answer:

x <= -5

Step-by-step explanation:

4x + 12 <= -8

4x <= -20

x <= -5

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The probability of flu symptoms for a person not receiving any treatment is 0.038. In a clinical trial of a common drug used to
alexgriva [62]

Answer:

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 1164, p = 0.038

So

\mu = E(X) = np = 1164*0.038 = 44.232

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 6.5231

Estimate the probability that at least 47 people experience flu symptoms.

Using continuity correction, this is P(X \geq 47 - 0.5) = P(X \geq 46.5), which is 1 subtracted by the pvalue of Z when X = 46.5.

Z = \frac{X - \mu}{\sigma}

Z = \frac{46.5 - 44.232}{6.5231}

Z = 0.35

Z = 0.35 has a 0.6368

1 - 0.6368 = 0.3632

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.

6 0
2 years ago
Jalen is buying a new car. He has the options below.
EleoNora [17]
The total number of choices Jalen has is six
7 0
3 years ago
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Jason’s credit card has an apr of 17.02% and a 30-day billling cycle. the following table details jason’s transactions with that
forsale [732]

The method of computing that would result in a greater finance charge is  a. the daily balance method will have a finance charge $1.02 greater than the adjusted balance method.

<h3>What is the Adjusted Balance Method?</h3>

This refers to the method of accounting that makes use of the owed amount of money at the end of a billing cycle to make its computation on an account after the credits are calculated.

Hence, we can see that when comparing the adjusted balance method to the daily balance method that calculates the interest charges at the end of the day, the daily balance method would have a higher finance charge.

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5 0
1 year ago
ASAP!! Please help me. I will not accept nonsense answers, but will mark as BRAINLIEST if you answer is correctly with solutions
Georgia [21]

Answer:

i feel like correct answrr would be C. Hope this helps!!!

7 0
2 years ago
The score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.
Artemon [7]

Answer:

P(X \geq 74) = 0.3707

Step-by-step explanation:

We are given that the score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Let X = Score of golfers

So, X ~ N(\mu=73,\sigma^{2}=3^{2})

The z score probability distribution is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = population mean = 73

           \sigma = standard deviation = 3

So, the probability that the score of golfer is at least 74 is given by = P(X \geq 74)

 P(X \geq 74) = P( \frac{X-\mu}{\sigma} \geq \frac{74-73}{3} ) = P(Z \geq 0.33) = 1 - P(Z < 0.33)

                                               =  1 - 0.62930 = 0.3707                  

Therefore, the probability that the score of golfer is at least 74 is 0.3707 .

3 0
3 years ago
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