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3 years ago

An instructor gives her class the choice to do 7 questions out of the 10 on an exam.

Mathematics

1 answer:

Maksim231197 [3]3 years ago

6 0

Answer:

(a) 120 choices

(b) 110 choices

Step-by-step explanation:

The number of ways in which we can select k element from a group n elements is given by:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways in which a student can select the 7 questions from the 10 questions is calculated as:

$10C7=\frac{10!}{7!(10-7)!}=120$

Then each student have 120 possible choices.

On the other hand, if a student must answer at least 3 of the first 5 questions, we have the following cases:

1. A student select 3 questions from the first 5 questions and 4 questions from the last 5 questions. It means that the number of choices is given by:

$(5C3)(5C4)=\frac{5!}{3!(5-3)!}*\frac{5!}{4!(5-4)!}=50$

2. A student select 4 questions from the first 5 questions and 3 questions from the last 5 questions. It means that the number of choices is given by:

$(5C4)(5C3)=\frac{5!}{4!(5-4)!}*\frac{5!}{3!(5-3)!}=50$

3. A student select 5 questions from the first 5 questions and 2 questions from the last 5 questions. It means that the number of choices is given by:

$(5C5)(5C2)=\frac{5!}{5!(5-5)!}*\frac{5!}{2!(5-2)!}=10$

So, if a student must answer at least 3 of the first 5 questions, he/she have 110 choices. It is calculated as:

50 + 50 + 10 = 110

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Answer:

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Step-by-step explanation:

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3 years ago

A school is organizing a weekend trip to a nature preserve. For each student, there is a $50 charge, which covers food and lodgi

k0ka [10]

The number of students going on the trip is 50.

Given,

A school is organizing a weekend trip to a nature preserve.

For each student, there is a $50 charge, which covers food and lodging. There is also a $25 charge per student for the bus.

The school must also pay a $15 cleaning fee for the bus.

The total cost of the weekend is $3,765.

We need to find out how many students will be going on the trip.

<h3>How do we find the number of items from the cost of each item and the total amount?</h3>

We simply divide the total cost by the cost of each item.

Example:

Total amount = 10

Each item cost = 2

Number of items = 10/2 = 5

Find the charge of each student.

= Food and lodging charge + bus charge

= $50 + $25

= $75

Find the additional charge for the trip.

= Cleaning charge

= $15

Find the total cost of the trip.

= $3,765

Find the total charge from the students.

= $3,765 - $15

= $3,750

Find the number of students who went on the trip.

= Total charge from the student's ÷ cost of each student

= 3750 ÷ 75

= 3750 / 75

= 50

This means 50 students.

Thus the number of students who went on the trip is 50.

Learn more about finding the number of students who like both games with a ratio given here:

brainly.com/question/24125573

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8 0

1 year ago

Read 2 more answers

The complete factorization of 12x² + 4x - 8 is:

PolarNik [594]

Answer:

D is the correct answer.

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2 years ago

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% 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 nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% 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 nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

2 years ago

Evaluate x - 20 when x=3

ryzh [129]

Answer:

-17

Step-by-step explanation:

Plug in 3:

3 - 20 =

Solve:

3-20 = -17

5 0

2 years ago

Read 2 more answers