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MrRissso [65]
3 years ago
6

The average value of y=v(x) equals 4 for 1≤x≤6, and equals 5 for 6≤x≤8. what is the average value of v(x) for 1≤x≤8 ?

Mathematics
1 answer:
Tomtit [17]3 years ago
7 0
The average value over the interval is the area under the curve divided by the width of the interval.

Area = 4*(6 -1) +5(8 -6) = 30
Width = 8 - 1 = 7

Average value = 30/7 = 4 2/7

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I’m so confused! Can someone help me with this? Please and thank you!
ludmilkaskok [199]

Answer:

D.

Step-by-step explanation:

1) the domain is x∈[0;+∞);

2) the solution is:

\sqrt{x}\leq 10; \ => \ x\leq 100;

3) finally, the solution with the domain is:

x∈[0;100].

7 0
2 years ago
Diamonds have a density of 3.5 LaTeX: \frac{g}{cm^3}g c m 3. How big is a diamond that has a mass of 0.10 g?
liubo4ka [24]

For this case we have that by definition, the density is given by:

d = \frac {M} {V}

Where:

M: It is the mass of the diamond

V: It is the volume of the diamond

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d = 3.5 \frac {g} {cm ^ 3}\\M = 0.10 \ g

So the volume is:

V = \frac {M} {d}\\V = \frac {0.10 \ g} {3.5 \frac {g} {cm ^ 3}}\\V = 0.02857\\V = 0.03 \ cm^3

Thus, the volume of the diamond is approximately 0.03 \ cm ^ 3

Answer:

0.03 \ cm ^ 3

3 0
3 years ago
30 POINTS PLEASE HELP
user100 [1]

Answer:

If all other factors are controlled, and the only variable being changed is the new medication, then Byron can conclude that the new medication is effective at reducing coughing. This can be further tested by repeated experiments by peers. 

Step-by-step explanation:

8 0
2 years ago
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Josiah used the distributive property to simplify the multiplication problem 9 × 4.7. Which of the following statements is true
zaharov [31]

Answer:

E. he multiplied 9 and 0.7 incorrectly.

Step-by-step explanation:

3 0
3 years ago
A. The lengths of pregnancies in a small rural village are normally distributed with a mean of 266 days and a standard deviation
Darya [45]

Step-by-step explanation:

a.

mean = 266

sd = 14

cumulative probability = 0.01 so the standard score = -2.33 and 2.33 to the right and left

we find X-upper and X-lower

X-lower = 266-2.33*14 = 233.38

X-upper = 266+2.33*14 = 298.62

Between 233.38 and 298.62

we have sample size = 35

X-lower = 266-2.33*14/√35 = 260.49

X-upper = 266+2.33*14/√35 = 271.5

Between 260.49 and 271.5

b. cumulative probaility = 0.25

standard score = 1.96 to the right and left

x-lower = 6.9-1.96x0.9 = 5.14

x-upper = 6.9+1.96x0.9 = 8.66

Between 5.14 and 8.66

if sample size = 45

x-lower = 6.9-1.96*0.9/√45 = 6.64

x-upper = 6.9+1.96*0.9/√45 = 7.2

Between 6.64 and 7.2

c. standard scores would have cut off value at 0.67 and -0,67

x-lower = 265.3-0.67x15.2 = 255.12

x-upper = 265.3+0.67x15.2 = 275.48

Between 255.12 and 275.48

d. we will have critical values at 1.00 and -1.00

X-lower = 265-1x16 = 249

x-upper = 265+1x16 = 281

Between 249 and 281

with sample size = 44

x-lower = 265-1x16/√44 = 262.59

x-upper = 265+1x16/√44 = 267.41

Between 262.59 and 267.41

3 0
3 years ago
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