Jeffrey is playing a game with the two spinners below in the attachment:

Consider a political discussion group consisting of 5 Democrats, 8 Republicans, and 5 Independents. Suppose that two group mem

Levart [38]

Answer:

4/9 and 5/18

Step-by-step explanation:

Total: 5+8+5=18

the probability of selecting a Republican is 8/18=4/9

the probability of selecting a Democrat is 5/18

6 0

4 years ago

• An equation written as Ax + By = C, where A, B, and C are real numbers is said to be in the

djyliett [7]

Answer:

standard form

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

7 0

4 years ago

2 to 6 ratio simplest form

aalyn [17]

The answer is 1:3. You have to divide by the greatest number possible. In this case it is 2. So you divide 2 by 2, and 6 by 2.

7 0

4 years ago

What is the answer to this

garik1379 [7]

When you subtract from a negative its more like adding - but keeping the negative sign.

so 35+1
36

keep the negative!
-36 :)

5 0

3 years ago

Read 2 more answers

Here are the weights, in pounds, of a sample of 13 adult female golden retriever dogs: 59.0, 54.1, 53.7, 51.6, 57.5, 58.7, 58.0,

lidiya [134]

Answer:

95% confidence interval for the population mean weight of adult female golden retrievers is [53.01 pounds , 56.78 pounds].

Step-by-step explanation:

We are given that the weights, in pounds, of a sample of 13 adult female golden retriever dogs :

59.0, 54.1, 53.7, 51.6, 57.5, 58.7, 58.0, 53.8, 48.9, 53.9, 51.6, 55.9, and 57.4.

Firstly, the pivotal quantity for 95% 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 weight = $\frac{\sum X }{n}$ = 54.9 pounds

s = sample standard deviation = $\frac{\sum (X-\bar X)^{2} }{n-1}$ = 3.12 pounds

n = sample of female = 13

$\mu$ = population mean weight

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.179 < $t_1_2$ < 2.179) = 0.95 {As the critical value of t at 12 degree

of freedom are -2.179 & 2.179 with P = 2.5%}

P(-2.179 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 2.179) = 0.95

P( $-2.179 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $2.179 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.179 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X +2.179 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = $\bar X \pm 2.179 \times {\frac{s}{\sqrt{n} } }$

= [ $54.9 - 2.179 \times {\frac{3.12}{\sqrt{13} } }$ , $54.9 +2.179 \times {\frac{3.12}{\sqrt{13} } }$ ]

= [53.01 pounds , 56.78 pounds]

Therefore, 95% confidence interval for the population mean weight of adult female golden retrievers is [53.01 pounds , 56.78 pounds].

3 0

4 years ago