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Vesnalui
1 month ago
6

The distance between two piers on a river is 45 miles. It takes a motorboat, moving downriver, 2 hr 15 min to complete the journ

ey. How much time does it take the same boat to go back upstream if the current flows at 2.5 mph?
Mathematics
1 answer:
PIT_PIT [12.4K]1 month ago
3 0
The duration is 3 hours. We can define a variable for our answer (x). Now we need to determine the boat's speed in still water (y). We can establish this equation: d/t = s: 45/(y + 2.5) = 2 1/4, leading us to find y = 17.5. Substituting in the current’s speed gives us y - 2.5. So, 17.5 - 2.5 is 15. By solving for x, we get 45/15 = 3, which gives us the final answer of 3 hours.
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Which of the following functions best describes this graph
lawyer [12517]

Response: B

Detailed explanation: This represents a quadratic function due to its parabolic shape, and it intersects the y-axis at 1.

7 0
2 months ago
The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop
lawyer [12517]

Answer:

a. \frac{1}{15}

b. \frac{2}{5}

c. \frac{14}{15}

d. \frac{8}{15}

Step-by-step explanation:

There are four desktop computers and two laptops.

On a specific day, we will set up 2 of these computers.

To find:

a. What is the probability that both selected computers are laptops?

b. What is the probability that both computers are desktops?

c. What is the probability that at least one computer is a desktop?

d. What is the probability that at least one of each type of computer is included?

Solution:

Using the probability formula for event E:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

a. The number of successful outcomes for both computers being laptops = _2C_2 = 1

Total possible outcomes = 15

The needed probability is \frac{1}{15}.

b. The successful outcomes for both being desktop computers = _4C_2=6

Total possible outcomes = 15

The required probability is \frac{6}{15} = \frac{2}{5}.

c. For at least one desktop:

Two scenarios exist:

1. 1 desktop and 1 laptop:

Successful outcomes = _2C_1\times _4C_1 = 8

2. Both are desktops:

Successful outcomes = _4C_2=6

Total successful outcomes = 8 + 6 = 14

The needed probability is \frac{14}{15}.

d. 1 desktop and 1 laptop:

Successful outcomes = _2C_1\times _4C_1 = 8

Total outcomes = 15

The required probability is \frac{8}{15}.

8 0
3 months ago
A baby wriggled so much that weighing him at the clinic was a problem. So the doctor held the baby and stood on a scale. Then th
babunello [11817]
Let X denote <span>the baby’s weight.
Let y represent </span>the doctor’s weight.
Let z stand for the nurse’s weight.

From the equations: x + y = 78 implies y = 78 - x
and x + z = 69 gives z = 69 - x
then we have x + y + z = 142

Substituting y = 78 - x and z = 69 - x into the equation x + y + z = 142

x + y + z = 142
x + 78 - x + 69 - x = 142
-x + 147 = 142
-x = -5
x = 5

Conclusion

<span>the baby’s weight was 5 kg</span>
6 0
2 months ago
Read 2 more answers
Circle 1 has center (−4, −7) and a radius of 12 cm. Circle 2 has center (3, 4) and a radius of 15 cm.
Svet_ta [12734]
Answer with explanation: Given that Circle 1 has a center at (−4, −7) and a radius of 12 cm, while Circle 2's center is at (3, 4) with a radius of 15 cm. Two circles are similar if one can be transformed and scaled to fit over the other, creating identical circles. The circles qualify as similar because the transformation rule (x,y) → (x+7,y+11) can be applied to Circle 1, followed by dilation using a scale factor of 5/4. Since Circle 1's center is at (-4,-7), we translate it to (3,4) through (-4+7,-7+11). With Circle 1 having a radius of 12 and Circle 2 having 15 cm, we denote the scale factor as k.
7 0
2 months ago
Read 2 more answers
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