C equals 2 pie r can you solve for r

Please someone help, give the right answer it’s important

Flura [38]

Answer:

2

Step-by-step explanation:

∠ MNQ = ∠ MNP + ∠ QNP

Since NP bisects ∠ MNQ , then

∠ MNP = ∠ QNP

3 0

2 years ago

Simplify 3/5-6i<br> A.15+18i/61<br> B.-15+18i/61<br> C.3i/6+5i<br> D.15+18i/-11

GrogVix [38]

Answer:

15/61+18/61i

Step-by-step explanation:

6 0

3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

<em>Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

<u>98% confidence interval for </u> $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago

In a chess variant, a "lord" can move one space at a time, either upward, or to the right, or diagonally up and to the right. ho

valentina_108 [34]

There are 48639 ways for a lord to move from the bottom left to top right corner of the 8 by 8 chessboard

<h3>Further explanation</h3>

The probability of an event is defined as the possibility of an event occurring against sample space.

$\large { \boxed {P(A) = \frac{\text{Number of Favorable Outcomes to A}}{\text {Total Number of Outcomes}} } }$

<h2>Permutation ( Arrangement )</h2>

Permutation is the number of ways to arrange objects.

$\large {\boxed {^nP_r = \frac{n!}{(n - r)!} } }$

<h2>Combination ( Selection )</h2>

Combination is the number of ways to select objects.

$\large {\boxed {^nC_r = \frac{n!}{r! (n - r)!} } }$

Let us tackle the problem.

This problem is similar to Pascal Triangle.

The Lord can move to the right or up in one way only.

We can put "1" in each box on the bottom and left to represent 1 possible movement.

To get the possibility of movement to another box, we will add the possibility of movement to the left, right, and diagonal squares as shown in Figure 1.

This process is repeated for the next box as shown in Figure 2.

Finally, after this process is repeated until the top right box, we get the results as shown in Figure 3.

There are 48639 ways for a lord to move from the bottom left to top right corner of the 8 by 8 chessboard.

<h3>Learn more</h3>