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Arlecino [84]
3 years ago
12

Need help with this problem asap

Mathematics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

plug in the day to the equation:

November 1st is day 305, so November 24th would be day 328

y = 116.65 sin(0.02(328) +1.89) + 731.71

= 116.65 sin(8.45) + 731.71

= 828.247 minutes of daylight

≈ 13.8 hours or 13:48 (because .8 of an hour is ≈ 48 minutes)

You might be interested in
29 yd 2ft 11 in + 55 yd 1 ft 10in + 13 yd 1ft 3in=
alexandr402 [8]
Add up all.

Yards = 29 + 55 + 13 = 97

Feet = 2 + 1 + 1 = 4

Inches = 11 + 10 + 3 = 24

In conclusion your final anwser is

97 yd. 4 ft. 24 in.

Hope this helps!
Have a great day.
7 0
3 years ago
Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 students. One
Naily [24]

Answer:

The expected value of X is E(X)=\frac{2754}{73} \approx 37.73 and the variance of X is Var(X)=\frac{226192}{5329} \approx 42.45

The expected value of Y is E(Y)=\frac{73}{2} \approx 36.5 and the  variance of Y is Var(Y)=\frac{179}{4} \approx 44.75

Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and  probability mass function p(x). The expected value, denoted by E(X) or \mu_x, is

E(X)=\sum_{x\in D} x\cdot p(x)

The probability mass function p_{X}(x) of X is given by

p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function p_{Y}(x) of Y is given by

p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}

The expected value of X is

E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)

E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73

The expected value of Y is

E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)

E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2

The variance of X is

E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)

E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}

Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45

The variance of Y is

E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)

E(Y^2)=28^2\cdot \frac{1}{4}+32^2\cdot \frac{1}{4} +42^2\cdot \frac{1}{4} +44^2 \cdot \frac{1}{4}\\\\E(Y^2)=196+256+441+484\\\\E(Y^2)=1377

Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75

8 0
3 years ago
How
Alecsey [184]

Answer:

A one solution

Step-by-step explanation:

4(x - 5) = 3x + 7

Distribute

4x - 20 = 3x+7

Subtract 3x from each side

4x-3x-20 = 3x+7-3x

x -20 = 7

Add 20 to each side

x -20+20 = 7+20

x = 27

There is one solution

5 0
3 years ago
Read 2 more answers
Please help me with thus pls
sergiy2304 [10]

Answer: 7

Step-by-step explanation:

8 0
3 years ago
What is the length of line segment PQ?<br><br> 4 units<br> 5 units<br> 6 units<br> 9 units
makkiz [27]
Hello! 

I was looking for this answer as well, but I took the quiz and got it correct. The answer is 5 units. 

good luck! 
3 0
3 years ago
Read 2 more answers
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