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3 years ago

PLEASE HELP ME evaluate cot pi/5

Mathematics

2 answers:

BaLLatris [955]3 years ago

6 0

0.628 is your answer. hope this helps

ICE Princess25 [194]3 years ago

4 0

1.376381921.37638192 is the answer

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I need help with this , I understand the thingy but I forgot what to do to solve it

True [87]

Answer:

if the team keeps winning at this rate they will have one 20 out of 25 games. The newspaper projected 23 out of 25 games. there's a three-game difference between The trend of the team and the prediction of the newspaper. The difference is 3 out of 25. divide 3 by 25 and you get 0.12 = 12%

4 0

3 years ago

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago

Please help! A is not -3 as I’ve seen some answer with that.

Arte-miy333 [17]

Answer:

a) x = 5

b) x = 4

Step-by-step explanation:

a) $\frac{3x+4}{2}$ = 9.5

3x+4 = 9.5(2)

3x = 19-4

x = 15/3

x = 5

b) $\frac{7+2x}{3}$ = 5

7+2x = 5(3)

2x = 15-7

x = 8/2

x=4

6 0

2 years ago

What is the property for 16=16

mestny [16]

Answer:

Reflexive property

Step-by-step explanation:

The property means that a real number is always equal to itself. For example, x=x. It works for all real numbers and is pretty simple. 3=3, 7.5=7.5 and etc

3 0

3 years ago

(i) Represent these two sets of data by a back-to-back stem-and-leaf diagram.

alexgriva [62]

<h3>Answer: </h3>

${\begin{tabular}{lll}\begin{array}{r|c|l}\text{Leaf (Ali)} & \text{Stem} & \text{Leaf (Kumar)}\\\cline{1-3} 7 & 4 & 1\ 2\ 3\ 6\ 6\ 9\ 9 \\ 9\ 8 & 5 & 2\ 2\ 3\\ 5\ 5 & 6 & \\ 7\ 2\ 0 & 7 & 8\ 8\ 9\\ 9\ 9\ 8\ 4\ 3\ 3\ 3\ 1\ 1 & 8 & 2\ 2\ 4\ 5\\ 9\ 8\ 1 & 9 & 0\ 2\ 5\\ \end{array} \\\\ \fbox{\text{Key: 7} \big| \text{4} \big| \text{1 means 4.7 for Ali and 4.1 for Kumar}} \end{tabular}}$

=========================================================

Explanation:

The data set for Ali is

8.3, 5.9, 8.3, 8.9, 7.7, 7.2, 8.1, 9.1, 9.8, 5.8,

8.3, 4.7, 7.0, 6.5, 6.5, 8.4, 8.8, 8.1, 8.9, 9.9

which when on a single line looks like this

8.3, 5.9, 8.3, 8.9, 7.7, 7.2, 8.1, 9.1, 9.8, 5.8, 8.3, 4.7, 7.0, 6.5, 6.5, 8.4, 8.8, 8.1, 8.9, 9.9

Let's sort the values from smallest to largest

4.7, 5.8, 5.9, 6.5, 6.5, 7.0, 7.2, 7.7, 8.1, 8.1, 8.3, 8.3, 8.3, 8.4, 8.8, 8.9, 8.9, 9.1, 9.8, 9.9

Now lets break the data up into separate rows such that each time we get to a new units value, we move to another row

4.7

5.8, 5.9

6.5, 6.5

7.0, 7.2, 7.7

8.1, 8.1, 8.3, 8.3, 8.3, 8.4, 8.8, 8.9, 8.9

9.1, 9.8, 9.9

We have these stems: 4, 5, 6, 7, 8, 9 which represent the units digit of the values. The leaf values are the tenths decimal place.

For example, a number like 4.7 has a stem of 4 and leaf of 7 (as indicated by the key below)

This is what the stem-and-leaf plot looks like for Ali's data only

$\ \ \ \ \ \ \ \ \text{Ali's data set}\\\\{\begin{tabular}{ll}\begin{array}{r|l}\text{Stem} & \text{Leaf}\\ \cline{1-2}4 & 7 \\ 5 & 8\ 9 \\ 6 & 5\ 5 \\ 7 & 0\ 2\ 7 \\ 8 & 1\ 1\ 3\ 3\ 3\ 4\ 8\ 9\ 9 \\ 9 & 1\ 8\ 9\\ \end{array} \\\\ \fbox{\text{Key: 4} \big| \text{7 means 4.7}} \\ \end{tabular}}$

The stem-and-leaf plot condenses things by tossing out repeated elements. Instead of writing 8.1, 8.1, 8.3 for instance, we can just write a stem of 8 and then list the individual leaves 1, 1 and 3. We save ourselves from having to write two more copies of '8'

Through similar steps, this is what the stem-and-leaf plot looks like for Kumar's data set only

$\ \ \ \ \ \ \ \ \text{Kumar's data set}\\\\{\begin{tabular}{ll}\begin{array}{r|l}\text{Stem} & \text{Leaf}\\ \cline{1-2}4 & 1\ 2\ 3\ 6\ 6\ 9\ 9 \\ 5 & \ 2\ 2\ 3\ \ \ \ \\ 6 & \\ 7 & 8\ 8\ 9 \\ 8 & 2\ 2\ 4\ 5\\ 9 & 0\ 2\ 5\\ \end{array} \\\\ \fbox{\text{Key: 4} \big| \text{1 means 4.1}} \\ \end{tabular}}$

Kumar doesn't have any leaves for the stem 6, so we will have that section blank. It's important to have this stem so it aligns with Ali's stem plot.

Notice that both stem plots involve the same exact set of stems (4 through 9 inclusive).

What we can do is combine those two plots into one single diagram like this

${\begin{tabular}{lll}\begin{array}{r|c|l}\text{Leaf (Ali)} & \text{Stem} & \text{Leaf (Kumar)}\\\cline{1-3} 7 & 4 & 1\ 2\ 3\ 6\ 6\ 9\ 9 \\ 8\ 9 & 5 & 2\ 2\ 3\\ 5\ 5 & 6 & \\ 0\ 2\ 7 & 7 & 8\ 8\ 9\\ 1\ 1\ 3\ 3\ 3\ 4\ 8\ 9\ 9 & 8 & 2\ 2\ 4\ 5\\ 1\ 8\ 9 & 9 & 0\ 2\ 5\\ \end{array} \\ \end{tabular}}$

Then the last thing to do is reverse each set of leaves for Ali (handle each row separately). The reason for this is so that each row of leaf values increases as you further move away from the stem. This is simply a style choice. This is somewhat similar to a number line, except negative values aren't involved here.

This is what the final answer would look like

The fact that Ali is on the left side vs Kumar on the right, doesn't really matter. We could swap the two positions and end up with the same basic table. I placed Ali on the left because her data set is on the top row of the original table given.

The thing you need to watch out for is that joining the stem and leaf for Ali means you'll have to read from right to left (as opposed to left to right). Always start with the stem. That's one potential drawback to a back-to-back stem-and-leaf plot. The advantage is that it helps us compare the two data sets fairly quickly.

6 0

2 years ago