Justin is looking to purchase a mountain bike from Bikes & More, WalMart, or K-Mart. Each store sells

Freshmen and Sophomores Solve by setting up a system of linear equations with 2 variables and 2 unknowns. A total of 1,595 first

VMariaS [17]

Let S = the sophomores
Let R = the freshmen
Short Answer C
s + r = 1595 (1)
s + 15 = r (2)

Substitute for r from equation (2) into equation (1)
s + s + 15 = 1595 Combine like terms
2s + 15 = 1595 Subtract 15 from both sides
2s = 1595 - 15
2s = 1580 Divide by 2
s = 1580/2
s = 795 sophmores.

s + r = 1595
795 + r = 1595 Subtract 795 from both sides
r = 1594 - 795
r = 800 Freshmen

Answer C <<<<<<

8 0

2 years ago

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Kevin wanted to go snowboarding for his vacation. Explain how he could make his decision regarding whether to go to Resort A or

Tcecarenko [31]

Answer:

Resort A has more consistent snowfall, so it shows less variation. However, the snowfall for Resort B has a higher median, and the interquartile range is higher (not larger), so it is more likely that Kevin will find a good snowfall at Resort B.

Thanks:) I just did it edg

Step-by-step explanation:

5 0

2 years ago

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The average of your first 6 tests is 82. If your test average after your seventh test is an 80.5, what did you score on the seve

mart [117]

So average is of the first 6 tests is 82 well that means

(test1 + test2+ test3+ test4 + test5+test6)/6 = 82

so now let do some cross multiplliying

test1 + test2 + test3 + test 4 + test5 +test6 = 82*6

test1 + test2 + test3 + test 4 + test5 +test6 = 492

now lets see if we can find that 7th test score

(test1 + test2 + test3 + test4 + test5 +test6 + test7)/7 = 80.5

So look we found test1 + test2 + test3 + test4 + test5 +test6 to be 492 so lets substitute.

(492 + test7 )/7 = 80.5

test7 = (80.5*7)-492 = 71.5

7 0

2 years ago

Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel time, is exponentially distributed,

seropon [69]

Answer and Step-by-step explanation:

Data provided in the question

Mean = 1.1 hours per call =

R = Mean rate = 1.6 per eight hour day

$\mu$ = $\frac{8}{1.6}$ = 5 per day

Based on the above information

a. The average number of customers is

$= \frac{R^2}{\mu(\mu- R)}$

$= \frac{1.6^2}{5(5- 1.6)}$

= 151

b. The system utilization is

$= \frac{R}{\mu}$

= $\frac{1.6}{5}$

= 0.32

c. The amount of time required is

= 1 - system utilization

= 1 - 0.32

= 0.68

And, there is 8 hours per day

So, it would be

= $0.68 \times 8$

= 5.44 hours

d. Now the probability of two or more customers is

$= 1 - (0.68 + 0.68 \times 0.32)$

= 0.1024

Therefore we simply applied the above formulas

4 0

2 years ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago