HELLO can some one please HELP ME OUT !!!

Based on the data in the two-way table, the probability of being 25-35 years and having a hemoglobin level above 11 is __ a.29 b

Alina [70]

The probability of being 25-35 years and having a haemoglobin level above 11 is 34%.

The probability of having a haemoglobin level above 11 is 36%.

Being 25-35 years and having a hemoglobin level above 11 are not dependent on each other.

<h3>What are the probabilities?
</h3>

Probability determines the odds that a random event would occur. The odds of the event happening lie between 0 and 1.

The probability of being 25-35 years and having a haemoglobin level above 11 = number of people between 25 - 35 that have a level above 11 / total number of people between 25 - 35

44 / 128 = 34%

The probability of having a haemoglobin level above 11 = number of people with a level above 11 / total number of respondents

153 / 429 = 36%

To learn more about probability, please check: brainly.com/question/13234031

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4 0

2 years ago

g A random sample of 100 students was taken. Eighty-five of the student in the sample experienced anxiety during the exam. We ar

BaLLatris [955]

Answer:

The test statistic is $z = 1.25$

Step-by-step explanation:

We are interested in determining whether or not the proportion of the student who experience anxiety during the exam is significantly more than 80%.

At the null hypothesis, we test if the proportion is 80%, that is:

$H_0: p = 0.8$

At the alternate hypothesis, we test if the proportion is more than 80%, that is:

$H_a: p > 0.8$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

80% is tested at the null hypothesis:

This means that $\mu = 0.8, \sigma = \sqrt{0.2*0.8} = 0.4$

A random sample of 100 students was taken. Eighty-five of the student in the sample experienced anxiety during the exam.

This means that $n = 100, X = \frac{85}{100} = 0.85$

The test statistic is

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.85 - 0.8}{\frac{0.4}{\sqrt{10}}}$

$z = 1.25$

The test statistic is $z = 1.25$

4 0

3 years ago

How do you do this problem?

erastovalidia [21]

Step-by-step explanation:

∫₀² x f(x²) dx

If u = x², then du = 2x dx, and ½ du = x dx.

When x = 0, u = 0. When x = 2, u = 4.

∫₀⁴ ½ f(u) du

½ (16)

8

7 0

2 years ago

Line segment SU is dilated to create S'U' using the dilation rule DQ,2.5. What is the distance, x, between points U' and U?

hodyreva [135]

Answer:

The answer is 6 units! Have a nice day! :D

Step-by-step explanation:

6 0

3 years ago

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The dot plots show rainfall total for several spring storms in highlands areas an lowlands areas what is the median rainfall for

tankabanditka [31]

*See attached picture showing the dot plots referred to

Answer:

16mm

Step-by-step Explanation:

Given the dot plot showing rainfall for Highlands storms, we can find out the median rainfall for Highlands storms by mere observing the dot plot and locating where the middle value falls within the data set that is represented on the dot plot.

Thus, we have 44 data sets that is being represented on the plot, with one dot representing one data set for rainfall. This means, since the total number of data set is even, our median lies between the 22nd and the 23rd dot or data set that is being plotted.

From the dot plot for Highlands, the 22nd and 23rd dots are 16 and 16 respectively.

Therefore, median rainfall for Highland storms = (16 + 16) ÷ 2 = 16.

8 0

2 years ago

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