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Leya [2.2K]
4 years ago
12

A pharmaceutical company collects data on the side effects participants report for a drug it is testing. The company finds that

the probability of experiencing a headache is 5 percent and the probability of experiencing both a headache and heartburn is 2 percent. If the probability of experiencing a headache or heartburn is 8 percent, what is the probability of experiencing heartburn?
Mathematics
2 answers:
Natasha2012 [34]4 years ago
6 0

Answer:

It's B

Step-by-step explanation:Hope this helps!

aniked [119]4 years ago
3 0
The answer is 5 percent , i just did it on edgen
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There are 350 to 450 students in Leah’s school. There are 1200 to 1300 students in her brother Zacks school. Choose a number of
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400 for Leah's school and 1250 for Zacks school
7 0
4 years ago
A certain type of thread is manufactured with a mean tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilogram
azamat

Answer:

(a) The variance decreases.

(b) The variance increases.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The standard deviation of sample mean is inversely proportional to the sample size, <em>n</em>.

So, if <em>n</em> increases then the standard deviation will decrease and vice-versa.

(a)

The sample size is increased from 64 to 196.

As mentioned above, if the sample size is increased then the standard deviation will decrease.

So, on increasing the value of <em>n</em> from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for <em>n</em> = 64 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7

The standard deviation of the sample mean for <em>n</em> = 196 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4

The standard deviation of the sample mean decreased from 0.7 to 0.4 when <em>n</em> is increased from 64 to 196.

Hence, the variance also decreases.

(b)

If the sample size is decreased then the standard deviation will increase.

So, on decreasing the value of <em>n</em> from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for <em>n</em> = 784 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2

The standard deviation of the sample mean for <em>n</em> = 49 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8

The standard deviation of the sample mean increased from 0.2 to 0.8 when <em>n</em> is decreased from 784 to 49.

Hence, the variance also increases.

6 0
3 years ago
What is the result of dividing x^3−7 by x + 2?
Pavlova-9 [17]

Answer:

Choice A: x^2−2x+4− 15/x+2

Step-by-step explanation:


5 0
4 years ago
The table shows a linear relationship between x and y.​ how can you tell that x and y are NOT porportianal
mote1985 [20]
If y/x does not always equal the same
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What is the slope of the line y=2/3x-5
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In y = mx + b form, the slope will be found in the m position

y = mx + b
y = (2/3)x - 5

so the slope(m) = 2/3
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3 years ago
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