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2 years ago

9

There are nine students in the chess club the club moderator select three of them to go to a chess tournament and how many ways

can she choose three of the nine students

2 answers:

Solnce55 [7]2 years ago

5 0

Answer:

84

Step-by-step explanation:

umka2103 [35]2 years ago

3 0

Answer:

apex: 84

Step-by-step explanation:

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I see that your new product is available for 412.50 on the website. How much is it if I buy it in the store? The website is a 25

OverLord2011 [107]

Answer with explanation:

Price of Product on the Website = $ 412.50

It is given that, product is offered at a discount of 25% on the website.

Let actual price of product on the store = $ x

Writing the above statement in terms of equation

→Price at store - Discount= Price at Website

$x-\frac{25\times x}{100}=412.50\\\\ \frac{75 x}{100}=412.50\\\\x=\frac{41250}{75}\\\\x=550$

Price at store of that Product = $ 550

3 0

2 years ago

When Ms. Sugar turned on her oven, the temperature inside was 70 degrees Fahrenheit. The temperature began to rise at a rate of

Stella [2.4K]

It would take fourteen minutes

7 0

2 years ago

Read 2 more answers

There are 135 people in a sport centre. 73 people use the gym. 59 people use the swimming pool. 31 people use the track. 19 peop

Anna71 [15]

Answer:

1/9

Step-by-step explanation:

135 in Sport centre: Total

59:swimming pool

31:track

19 both swimming and gym

16 gym and track

4 all three facilities

4 people use all three facilities, then

16 - 4 = 12 people use the gym and the track and do not use the pool;

9 - 4 = 5 people use the pool and the track and do not use the gym;

19 - 4 = 15 people use the gym and the pool and do not use the track.

At least two facilities use 4 + 12 + 5 + 15 = 36 people, 4 of them use all three facilities. Thus, the probability that a randomly selected person which uses at least two facilities, uses all the facilities is

4/36=1/9

Hope this helps!!!

6 0

2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population 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 Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

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

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago

each transformation is performed on the line with the equation y=2x-1. write the equation of the new line. vertical translation

Phoenix [80]

Answer:

y2= 2x-4

y3=6x-1

y4= x-1

y5=2x

Step-by-step explanation:

for y=2x-1

1) for a vertical translation down of 3 units

y2= y-3 =(2x-1)-3= 2x-4

y2= 2x-4

2) for a slope increased by 4

y3= y+ 4x = 2x-1 +4x = 6x-1

y3=6x-1

3) for sloped divided in half. slope of y : m=2 → slope of y4=2/2 =1

y4= x-1

4) shifted up (vertical translation) of 1 unit

y5= y+1 = 2x-1+1=2x

y5=2x

6 0

1 year ago