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Alexxx [7]
3 years ago
15

What is the probability that a junior non-Nutrition major and then a sophomore Nutrition major are chosen at random? Express you

r answer as a fraction or a decimal number rounded to four decimal places.

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
7 0

Answer:

0.0032

The complete question as seen in other website:

There are 111 students in a nutrition class. The instructor must choose two students at random Students in a Nutrition Class Nutrition majors Academic Year Freshmen non-Nutrition majors 17 18 Sophomores Juniors 13 Seniors 18 Copy Data. What is the probability that a senior Nutrition major and then a junior Nutrition major are chosen at random? Express your answer as a fraction or a decimal number rounded to four decimal places.

Step-by-step explanation:

Total number of in a nutrition class = 111 students

To determine the probability that the two students chosen at random is a junior non-Nutrition major and then a sophomore Nutrition major, we would find the probability of each of them.

Let the probability of choosing a junior non-Nutrition major = Pr (j non-N)

Pr (j non-N) = (number of junior non-Nutrition major)/(total number students in nutrition class)

There are 13 number of junior non-Nutrition major

Pr (j non-N) = 13/111

Let the probability of choosing a sophomore Nutrition major = Pr (S N-major)

Pr (S N-major)= (number of sophomore Nutrition major)/(total number students in nutrition class)

There are 3 number of sophomore Nutrition major

Pr (S N-major) = 3/111

The probability that the two students chosen at random is a junior non-Nutrition major and then a sophomore Nutrition major = 13/111 × 3/111

= 39/12321

= 0.0032

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blagie [28]

The answer is very simple .

First you need to determine the slope, which is the reciprocal of the first line.

Then,

y = mx + b

m => -1/m

The perperdicular line has a new slope equal to m = -1/2

In the formula (y - y1) = m ( x - x1 ) + b , you should sustitude this value.

So the final equation will be

-3 = -1/2 (0) + b

b = - 3

y = (-1/2)x - 3

That is the solution

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1 year ago
Find the value of x.<br><br> :]]
Mekhanik [1.2K]

Answer:

144

Step-by-step explanation:

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2 years ago
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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
A fair spinner has 10 equal sections there is a probability scale
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Answer:

1/10

Step-by-step explanation:

Because there is ten different numbers so therefore it would be 10/1

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let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep
Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

x_1,x_2 and x_3 are linearly independent.

By definition of linear independent there exits three scalar a_1,a_2 and a_3 such that

a_1x_1+a_2x_2+a_3x_3=0

Where a_1=a_2=a_3=0

y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1

We have to prove that y_1,y_2 and y_3 are linearly independent.

Let b_1,b_2 and b_3 such that

b_1y_1+b_2y_2+b_3y_3=0

b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0

b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0

(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0

b_1+b_3=0

b_1=-b_3...(1)

b_1+b_2=0

b_1=-b_2..(2)

b_2+b_3=0

b_2=-b_3..(3)

Because x_1,x_2 and x_3 are linearly independent.

From equation (1) and (3)

b_1=b_2...(4)

Adding equation (2) and (4)

2b_1==0

b_1=0

From equation (1) and (2)

b_3=0,b_2=0,b_3=0

Hence, y_1,y_2 and y_3 area linearly independent.

5 0
3 years ago
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