All categories

3 years ago

I need help with these pages please answer them all correctly this is not a freebie don't put random words or letters thanks.

Mathematics

1 answer:

Anna007 [38]3 years ago

7 0

Answer:

the question is: why should ppl answer your questions correctly and help you when you dont answer thier questions and help them when they need it? why help some one who wont help you?

You might be interested in

What is the percent of change from 50 to 23?

Taya2010 [7]

Answer:

50 is the old value and 23 is the new value. In this case we have a negative change (decrease) of -54 percent because the new value is smaller than the old value. Using this tool you can find the percent decrease for any value.

7 0

3 years ago

Read 2 more answers

Kevin usually walks 30km at x km/h for his walkathon training. If he increases his speed by another 1 km/h, he can complete the

Naily [24]

Answer:

x = 5km/h

Step-by-step explanation:

If he walks 5km/h, he would reach 30km in 6hrs. However, if he walks 1km/hr faster which is 6km/h, he wouls reach 30km in 5hrs. This meaning he can complete the same distance 1 hour faster. Hope this helps you. Follow me for faster replies. And smash that Thank you button.❤

7 0

3 years ago

Consider a sample with data values of 27, 24, 21, 16, 30, 33, 28, and 24. Compute the 20th, 25th, 65th, and 75th percentiles. 20

densk [106]

Answer:

$P_{20} = 20$ --- 20th percentile

$P_{25} = 21.75$ --- 25th percentile

$P_{65} = 27.85$ --- 65th percentile

$P_{75} = 29.5$ --- 75th percentile

Step-by-step explanation:

Given

27, 24, 21, 16, 30, 33, 28, and 24.

N = 8

First, arrange the data in ascending order:

Arranged data: 16, 21, 24, 24, 27, 28, 30, 33

Solving (a): The 20th percentile

This is calculated as:

$P_{20} = 20 * \frac{N +1}{100}$

$P_{20} = 20 * \frac{8 +1}{100}$

$P_{20} = 20 * \frac{9}{100}$

$P_{20} = \frac{20 * 9}{100}$

$P_{20} = \frac{180}{100}$

$P_{20} = 1.8th\ item$

This is then calculated as:

$P_{20} = 1st\ Item +0.8(2nd\ Item - 1st\ Item)$

$P_{20} = 16 + 0.8*(21 - 16)$

$P_{20} = 16 + 0.8*5$

$P_{20} = 16 + 4$

$P_{20} = 20$

Solving (b): The 25th percentile

This is calculated as:

$P_{25} = 25 * \frac{N +1}{100}$

$P_{25} = 25 * \frac{8 +1}{100}$

$P_{25} = 25 * \frac{9}{100}$

$P_{25} = \frac{25 * 9}{100}$

$P_{25} = \frac{225}{100}$

$P_{25} = 2.25\ th$

This is then calculated as:

$P_{25} = 2nd\ item + 0.25(3rd\ item-2nd\ item)$

$P_{25} = 21 + 0.25(24-21)$

$P_{25} = 21 + 0.25(3)$

$P_{25} = 21 + 0.75$

$P_{25} = 21.75$

Solving (c): The 65th percentile

This is calculated as:

$P_{65} = 65 * \frac{N +1}{100}$

$P_{65} = 65 * \frac{8 +1}{100}$

$P_{65} = 65 * \frac{9}{100}$

$P_{65} = \frac{65 * 9}{100}$

$P_{65} = \frac{585}{100}$

$P_{65} = 5.85\th$

This is then calculated as:

$P_{65} = 5th + 0.85(6th - 5th)$

$P_{65} = 27 + 0.85(28 - 27)$

$P_{65} = 27 + 0.85(1)$

$P_{65} = 27 + 0.85$

$P_{65} = 27.85$

Solving (d): The 75th percentile

This is calculated as:

$P_{75} = 75 * \frac{N +1}{100}$

$P_{75} = 75 * \frac{8 +1}{100}$

$P_{75} = 75 * \frac{9}{100}$

$P_{75} = \frac{75 * 9}{100}$

$P_{75} = \frac{675}{100}$

$P_{75} = 6.75th$

This is then calculated as:

$P_{75} = 6th + 0.75(7th - 6th)$

$P_{75} = 28 + 0.75(30- 28)$

$P_{75} = 28 + 0.75(2)$

$P_{75} = 28 + 1.5$

$P_{75} = 29.5$

7 0

3 years ago

Each of the following situations describes a normally distributed population for which a statistical procedure should be used. M

densk [106]

Answer:

(a) Correct option (1).

(b) Correct option (3).

Step-by-step explanation:

A one-sample z-test for mean is used to determine whether the claimed value of mean is significant or not. To perform this test we can also use the t-distribution.

The t-test is used if the following conditions are given:

The population is normally distributed
The population standard deviation is not known
The sample selected is large enough.

Similarly the (1 - α)% confidence interval for population mean can be computed either using the z-interval or the t-interval.

The same conditions should be applied to use the t-interval.

(a)

In the first case, Gale wants to determine the likely range of weight of whole coffee beans from her trusty two-tablespoon scoop to create a reliable method of brewing the perfect cup of coffee.

She selects a sample of n = 15 scoops and perform the test.

Since the population standard deviation is not known and Gale wants to know the range of weight of whole coffee beans , she should use a one sample t-confidence interval for a mean.

The correct option is (1).

(b)

In the second case, Tanya wants to determine whether the mean IQ score is at least 112 or not.

She selects a a random sample n = 20 composers and compute the statistic to perform the one-sample test.

Since the population standard deviation is not known, Tanya should use one sample t-test for a mean.

The correct option is (3).

(c)

In the second case, Jason wants to estimate what the balance of his retirement account will be in one month.

He selects the data from 1950 to 2015 and calculates this index has had an average monthly return of 0.69% with a standard deviation of 4.15%.

Now, the sample selected is quite large. The Central limit theorem can be used to approximate the sampling distribution of sample mean.

Then, $\bar X\sim N(\mu_{\bar x}, \sigma_{\bar x})$

So, Jason can use the one sample z-test for a mean to determine the balance of his retirement account will be in one month.

The correct option is (2).

7 0

4 years ago

Can someone please help me with this?

Levart [38]

Answer:

5 is the answer. hope this help you

5 0

3 years ago

Read 2 more answers