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kkurt [141]
3 years ago
13

On a certain hot summer's day, 346 people used the public swimming pool. The daily prices are $1.75 for children and $2.00 for a

dults. The
receipts for admission totaled $640.00. How many children and how many adults swam at the public pool that day?
Mathematics
1 answer:
marishachu [46]3 years ago
3 0
208 children and 138 adults
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Jessica ran on the track team at her school. She finished the 200-meter race in 54 seconds. What was jessica's average speed,in
Scrat [10]

Answer:

v =\frac{x}{t}

Where x represent the distance and t the time if we replace the info given we got:

v = \frac{200m}{54 s} =3.704 \frac{m}{s}

Then the average speed for this case in m/s is 3.704 for the 200 meter race

Step-by-step explanation:

The problem provides for this case the amount of distance travelled by Jessica in a given time. And we know that travels 200m in 54 seconds.

We want to find the average speed and we can use the following formula:

v =\frac{x}{t}

Where x represent the distance and t the time if we replace the info given we got:

v = \frac{200m}{54 s} =3.704 \frac{m}{s}

Then the average speed for this case in m/s is 3.704 for the 200 meter race

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4 years ago
7 out of 22 patients have a heart problem, 8 are randomly selected
beks73 [17]

Answer:6

Step-by-step explanation:

4 0
3 years ago
What is the slope of the line through 0,4 and 3,-2
zysi [14]

Answer: -2

Step-by-step explanation: slope is difference in y values over difference in x values

6 0
3 years ago
Differentiate x^(-3)/3-2x^2-x^(-1)
GaryK [48]

Answer:

-2x^2

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Simplify

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4 years ago
Two radar stations 5.5 miles apart are tracking an airplane. The straight line distance
Vesnalui [34]

Answer:

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Step-by-step explanation:

I have attached an image showing this elevation.

From the image, let's first find the angle A by using cosine rule.

Thus;

8.1² = 5.5² + 13.1² - 2(5.5 × 13.1)cos A

65.61 = 30.25 + 171.61 - 144.1cos A

144.1cos A = 171.61 + 30.25 - 65.61

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cosA = 136.25/144.1

cosA = 0.9455

A = cos^(-1) 0.9455

A = 19°

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3 years ago
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