Determine the coordinates of the vertices of the image of ABC after a reflection in the line y = -x.

a lottery game has balls numbered 1 through 21. what is the probability of selecting a ball with a prime number on it or a 10

Ronch [10]

Answer:

3/7

Step-by-step explanation:

The prime numbers that lie between 1 and 21 are 2,3,5,7,11,13,17, and 19. These 8 balls, plus the 10-ball comes to a total of 9 balls that could be picked out of the 21. Therefore, the probability of picking one of these balls is 9/21 or 3/7

6 0

4 years ago

Read 2 more answers

for the first 18 days of the month, buffet dinner generate $11,800 in sales. The buffet sale target for the month is $18,650. Gi

sesenic [268]

Your Daily Sale Goal should be $611.61 to achieve the $18.65k monthly goal

8 0

3 years ago

A rectangle has length 32 inches and its perimeter is 6 times its width. What is its width in inches

professor190 [17]

Answer:

16

Step-by-step explanation:

6x = 2(x + 32)

I hope this helps!

8 0

3 years ago

How many times can 43 go into 100

lara [203]

About 2 times at least it would also be 2.3 some other numbers also

6 0

3 years ago

for a class project, a political science student at a large university wants to estimate the percent of students that are regist

nekit [7.7K]

Answer:

90% confidence interval for the true percent of students that are registered voters is [0.56 , 0.64].

Step-by-step explanation:

We are given that for a class project, a political science student at a large university wants to estimate the percent of students that are registered voters.

He surveys 500 students and finds that 300 are registered voters.

Firstly, the pivotal quantity for 90% confidence interval for the true proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students who are registered voters = $\frac{300}{500}$ = 0.60

n = sample of students = 500

p = true percent of students

<em>Here for constructing 90% confidence interval we have used One-sample z proportion test statistics.</em>

So, 90% confidence interval for the true proportion, p is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90 {As the critical value of z at 5%

level of significance are -1.6449 & 1.6449}

P(-1.6449 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.6449) = 0.90

P( $-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> =[ $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.60-1.6449 \times {\sqrt{\frac{0.60(1-0.60)}{500} } }$ , $0.60+1.6449 \times {\sqrt{\frac{0.60(1-0.60)}{500} } }$ ]

= [0.56 , 0.64]

Therefore, 90% confidence interval for the true percent of students that are registered voters is [0.56 , 0.64].

Interpretation of the above confidence interval is that we are 90% confident that the true percent of students that are registered voters will lie between 0.56 and 0.64.

6 0

3 years ago