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3 years ago

Bryan worked 20 hours last week. He earned $100. How much did he earn per hour?

Mathematics

2 answers:

castortr0y [4]3 years ago

5 0

He earned $5 an hour

mixer [17]3 years ago

4 0

The answer would be 5 dollars an hour

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Select the curve generated by the parametric equations. Indicate with an arrow the direction in which the curve is traced as t i

bixtya [17]

Answer:

length of the curve = 8

Step-by-step explanation:

Given parametric equations are x = t + sin(t) and y = cos(t) and given interval is

−π ≤ t ≤ π

Given data the arrow the direction in which the curve is traces means

the length of the curve of the given parametric equations.

The formula of length of the curve is

$\int\limits^a_b {\sqrt{\frac{(dx}{dt}) ^{2}+(\frac{dy}{dt}) ^2 } } \, dx$

Given limits values are −π ≤ t ≤ π

x = t + sin(t) ...….. (1)

y = cos(t).......(2)

differentiating equation (1) with respective to 'x'

$\frac{dx}{dt} = 1+cost$

differentiating equation (2) with respective to 'y'

$\frac{dy}{dt} = -sint$

The length of curve is

$\int\limits^\pi_\pi {\sqrt{(1+cost)^{2}+(-sint)^2 } } \, dt$

$\int\limits^\pi_\pi \, {\sqrt{(1+cost)^{2}+2cost+(sint)^2 } } \, dt$

on simplification , we get

here using sin^2(t) +cos^2(t) =1 and after simplification , we get

$\int\limits^\pi_\pi \, {\sqrt{(2+2cost } } \, dt$

$\sqrt{2} \int\limits^\pi_\pi \, {\sqrt{(1+1cost } } \, dt$

again using formula, 1+cost = 2cos^2(t/2)

$\sqrt{2} \int\limits^\pi _\pi {\sqrt{2cos^2\frac{t}{2} } } \, dt$

Taking common $\sqrt{2}$ we get ,

$\sqrt{2}\sqrt{2} \int\limits^\pi _\pi ( {\sqrt{cos^2\frac{t}{2} } } \, dt$

$2(\int\limits^\pi _\pi {cos\frac{t}{2} } \, dt$

$2(\frac{sin(\frac{t}{2} }{\frac{t}{2} } )^{\pi } _{-\pi }$

length of curve = $4(sin(\frac{\pi }{2} )- sin(\frac{-\pi }{2} ))$

length of the curve is = 4(1+1) = 8

<u>conclusion</u>:-

The arrow of the direction or the length of curve = 8

7 0

3 years ago

How many terms are in the expression shown below? x^2 - 10xy + 3y + y^2 - 1

Viefleur [7K]

<span>There are 5 terms in the expression</span>

6 0

3 years ago

You are on vacation and want to buy a pair of sunglasses for $10 or less. You find a pair with a price tag

enyata [817]

I notice that The price is higher than the original price
And I wonder it is because of the tax?

5 0

3 years ago

True or False, ∠XZY and ∠PZQ are vertical angles, then ∠XZY≅∠PZQ

Natasha_Volkova [10]

Answer:

True

Step-by-step explanation:

If ∠XZY and ∠PZQ are vertical angles,

then ∠XZY≅∠PZQ

(vertical angles are congruent)

3 0

3 years ago

6) Supplementary Exercise 5.51

tresset_1 [31]

Answer:

$P(X \le 4) = 0.7373$

$P(x \le 15) = 0.0173$

$P(x > 20) = 0.4207$

$P(20\ge x \le 24)= 0.6129$

$P(x = 24) = 0.0236$

$P(x = 15) = 1.18\%$

Step-by-step explanation:

Given

$p = 80\% = 0.8$

The question illustrates binomial distribution and will be solved using:

$P(X = x) = ^nC_xp^x(1 - p)^{n-x}$

Solving (a):

Given

$n =5$

Required

$P(X\ge 4)$

This is calculated using

$P(X \le 4) = P(x = 4) +P(x=5)$

This gives:

$P(X \le 4) = ^5C_4 * (0.8)^4*(1 - 0.8)^{5-4} + ^5C_5*0.8^5*(1 - 0.8)^{5-5}$

$P(X \le 4) = 5 * (0.8)^4*(0.2)^1 + 1*0.8^5*(0.2)^0$

$P(X \le 4) = 0.4096 + 0.32768$

$P(X \le 4) = 0.73728$

$P(X \le 4) = 0.7373$ --- approximated

Solving (b):

Given

$n =25$

Required

$P(X\le 15)$

This is calculated as:

$P(X\le 15) = 1 - P(x>15)$ --- Complement rule

$P(x>15) = P(x=16) + P(x=17) + P(x =18) + P(x = 19) + P(x = 20) + P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)$

$P(x > 15) = {25}^C_{16} * p^{16}*(1-p)^{25-16} +{25}^C_{17} * p^{17}*(1-p)^{25-17} +{25}^C_{18} * p^{18}*(1-p)^{25-18} +{25}^C_{19} * p^{19}*(1-p)^{25-19} +{25}^C_{20} * p^{20}*(1-p)^{25-20} +{25}^C_{21} * p^{21}*(1-p)^{25-21} +{25}^C_{22} * p^{22}*(1-p)^{25-22} +{25}^C_{23} * p^{23}*(1-p)^{25-23} +{25}^C_{24} * p^{24}*(1-p)^{25-24} +{25}^C_{25} * p^{25}*(1-p)^{25-25}$

$P(x > 15) = 2042975 * 0.8^{16}*0.2^9 +1081575* 0.8^{17}*0.2^8 +480700 * 0.8^{18}*0.2^7 +177100 * 0.8^{19}*0.2^6 +53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0$