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2 years ago

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7. In a lottery, players must match five numbers plus a bonus number. Five white balls are chosen from 59 white balls numbered f

rom 1 to 59, and one red ball (the bonus number)
is chosen from 35 red balls numbered 1 to 35. How many different results are possible?

1 answer:

11111nata11111 [884]2 years ago

7 0

Answer:

There are 175,223,510 ways.

Step-by-step explanation:

If the order of the white balls doesnt metter, we have as many possibilities to pick white balls as ways of pick 5 elements from a set of 59. That number is the combinatorial number of 59 with 5 given by

${59 \choose 5} = \frac{59!}{(59-5)!5!} = 5,006,386$

We have 35 different ways to pick the red ball. Combining both events, we have

35*5,006,386 = 175,223,510

ways of picking the 5 numbers and the bonus one.

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Jordan had the following scores on her math tests last quarter: 96, 89, 79, 85, 87, 94,

Lisa [10]

Answer:

a) 5.5

b) None

Step-by-step explanation:

The given data set is {96,89,79,85,87,94,96,98}

First we must find the mean.

$\bar X=\frac{96+89+79+85+87+94+96+98}{8}=\frac{724}{8}=90.5$

We now find the absolute value of the distance of each value from the mean.

This is called the absolute deviation

{ $|96-90.5|,|89-90.5|,|79-90.5|,|85-90.5|,|87-90.5|,|94-90.5|,|96-90.5|,|98-90.5|$ }

{ $5.5,1.5,11.5,5.5,3.5,3.5,5.5,7.5$ }

We now find the mean of the absolute deviations

$MAD=\frac{5.5+1.5+11.5+5.5+3.5+3.5+5.5+7.5}{8} =\frac{44}{8} =5.5$

The least absolute deviation is 1.5. This is not within one absolute deviation.

Therefore none of the data set is closer than one mean absolute deviation away from the mean.

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2 years ago

Jennifer has a square bedroom. The area of her room is 144 ft2. What is the length of one side of Jennifer’s room?

Jobisdone [24]

Answer:

Option B. $12\ ft$

Step-by-step explanation:

we know that

The area of a square is equal to

$A=x^{2}$

where

x is the length side of the square

In this problem we have

$A=144\ ft^{2}$

substitute and solve for x

$144=x^{2}$

suqre root both sides

$x=12\ ft$

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2 years ago

300-297+294-291+288-285+...+6-3

vladimir2022 [97]

300 - 297 = 3

294 - 291 = 3

288 - 285 = 3

So, that is a sequence of sums of number 3: 3 + 3 + 3 + .... + 3

How may times will the number 3 be added?

Note that 300 is reduced in 6 units each time => 300 / 6 = 50 => 3 will be added 50 times.

=> 50 * 3 = 150

Answer: 150

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2 years ago

Tiesha enjoys reading in her spare time. She reads 4 pages every 1/10 of an hour.

dangina [55]

Answer:

p = 40h

Step-by-step explanation:

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2 years ago

Read 2 more answers

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%.

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2 years ago