Which of the following measurements is the most accurate for a pellet with a mass of 3.25 grams?

Does somebody knows how to do this?

MrMuchimi

Answer:

Area of the triangle = 6x3/2=18/2=9cm²

now find semicircle then add

area of the semicircle= πr²/2=π6²/2=36π/2=18π

now add and you get

Area of the region = (18π+9)cm²

Step-by-step explanation:

5 0

3 years ago

Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbic

jolli1 [7]

Answer:

The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

The randomly sampled 400 dogs from homes where an herbicide was used on a regular basis, diagnosing lymphoma in 230 of them.

This means that:

$p_h = \frac{230}{400} = 0.575, s_h = \sqrt{\frac{0.575*0.425}{400}} = 0.0247$

Of 200 dogs randomly sampled from homes where no herbicides were used, only 25 were found to have lymphoma.

This means that:

$p_n = \frac{25}{200} = 0.125, s_n = \sqrt{\frac{0.125*0.875}{200}} = 0.0234$

Distribution of the difference:

$p = p_h - p_n = 0.575 - 0.125 = 0.45$

$s = \sqrt{s_h^2+s_n^2} = \sqrt{0.0247^2 + 0.0234^2} = 0.034$

Confidence interval:

The confidence interval is:

$p \pm zs$

In which

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

95% confidence level

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 bound is $0.45 - 1.96(0.034) = 0.3834$

The upper bound is $0.45 + 1.96(0.034) = 0.5166$

The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

3 0

3 years ago

Use the line plot above to answer all 4 questions. Label your responses (a), (b), (c), and (d). When asked to explain you must u

mojhsa [17]

A) it is left skewed
B) the median is 5
C) the mean is 5.15
D) the mean would be more affected (a change of 1.05 versus a change of 0.5).

The majority of the data is to the right of the graph; this means it is left skewed.

To find the median, write all of the data values out:
2, 3, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7

The middle value is 5.

We find the sum of this set of values and divide by 13, the number of data points, to find the mean:
2+3+4+4+5+5+5+6+6+6+7+7+7 = 67/13 = 5.15

If we added an additional data value at 20, the new median would be 5.5. The new mean would be (67+20)/14 = 6.2. The mean changes more than the median.

4 0

3 years ago

Find the image of each polygon with the given vertices after a dilation centered at the origin with the given scale factor. J(2,

MAVERICK [17]

Applying a dilation centered at the origin with scale factor $k$ has the effect of multiplying both coordinates of a point by $k$ :