Can you helo please !!!!

A study shows that 78% of the population of all calculus students consider calculus an exciting subject. Suppose 20 students ar

alexdok [17]

Answer:

0.046145 = 4.6145%

Step-by-step explanation:

We need to find the probability of 19 or 20 (all) students considering calculus to be an exciting subject, so first we find the probability for 19 students, then for 20 students, and we sum these probabilities.

The probability of the student NOT considering calculus to be an exciting subject is 1 - 0.78 = 0.22

For 19 students, we also have to calculate a combination of 20 choose 19, as these 19 students can be in different "position" among the 20 students, so:

C(20,19) = 20!/19! = 20

Probability of 19 students considering calculus to be an exciting subject:

C(20,19) * 0.78^19 * 0.22^1 = 0.039196756

Probability of 20 students considering calculus to be an exciting subject:

C(20,20) * 0.78^20 = 1 * 0.78^20 = 0.006948516

The final probability is:

0.039196756 + 0.006948516 = 0.046145272 = 4.6145272%

Rounding to six decimal places: 0.046145 = 4.6145%

5 0

3 years ago

Solve by augmented matrix (GJ)<br> 2x + 3y – 4z = 1<br> 4x + y + 2z = -3<br> pls help!!

otez555 [7]

Answer:

3x-y-2z=4. 2x+3y-4z=1

4x-7y-6z=-7. -x+6y+4z=11

- +. +. +. x+9y=12

____________.

-x +6y +4z=11.

5 0

2 years ago

A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

8 0

3 years ago

PLS HELP ME WITH THIS !!

Mashutka [201]

Part 1:
180 = 5p+98 + 9p-2
180 = 14p + 96
84 = 14p
6 = p

part 2:
angle S will be the same as angle Q, so:
9p-2
9(6)-2
54-2
M

3 0

3 years ago

Mr. Whitley works at a furniture store and

photoshop1234 [79]

750 I iiiiiiiiiioiiii

4 0

3 years ago