Will works 7 hours a day, 6 days a week.

1 answer:

6 0

A day
= £9.20x7
= £64.40

6 days
£64.40x6= £386.40

The wages he has after he shared it with his mom
£386.40/7x5
= 55.20x5
= £276

£1932/£276= 7

It will take him 7 weeks to afford a car worth £1932.

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A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

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2 years ago

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A set of data has a mean of 56.1. The data follows a normal distribution curve and has a standard deviation of 8.2. Find the pro

Whitepunk [10]

Given that mean=56.1 and standard deviation=8.2, P(x>67.5) will be found as follows:
The z-score is given by:
z=(x-μ)/σ
thus the z-score will be given by:
z=(67.5-56.1)/8.2
z=11.4/8.2
z=1.39
thus
P(z=1.39)=0.9177
thus:
P(x>67.5)=1-P(z>0.9177)
=1-0.9177
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Answer: A. 0.0823

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Describe the graph represented by the equation r=5/(3+2 sin theta).

Yuki888 [10]

We have the following equation:

$r= \frac{5}{3+2sin(\theta)}$

If we graph this equation we realize that in fact this is an ellipse with major axis matching the y-axis. So we can recognize these characteristics:

1. Center of the ellipse:

The midpoint C of the line segment joining the foci is called the center of the ellipse. So in this exercise this point is as follows:

$C(0, -2)$

2. Length of major axis:

The line through the foci is called the major axis, so in the figure if you go from -5, at the y-coordinate, and walk through this major axis to the coordinate 1, the distance you run is the length of the major axis, that is:

$6 \ units$

3. Length of minor axis:

The line perpendicular to the foci through the center is called the minor axis. So in the figure if you go from -2, at the x-coordinate, and walk through this minor axis to the coordinate 2, the distance you run is the length of the minor axis, that is:

$4 \ units$

4. Foci:

Let's find c as follows:

$c=\sqrt{a^{2}-b^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}$

Then the foci are:

$f_{1}=(0, \sqrt{5}-2)$

$f_{2}=(0, -\sqrt{5}-2)$