Where to locate the nonnegative abscissa, positive ordinate

Ice cream usually comes in 1.5-quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some v

Troyanec [42]

Answer:

(a) The expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.

(b) The expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.

(c) Because the variance of each variable is dependent on the other.

Step-by-step explanation:

The random variable X and Y are defined as follows:

X = amount of ice cream in the box

Y = amount of ice cream scooped out

The information provided is:

E (X) = 48

SD (X) = 1

V (X) = 1

E (Y) = 2

SD (Y) = 0.25

V (Y) = 0.0625

(a)

The total amount of ice-cream served at the party can be expressed as:

X + 3Y.

Compute the expected value of (X + 3Y) as follows:

$E(X+3Y)=E(X)+3E(Y)\\= 48+(3\times2)\\=48+6\\=54$

Compute the variance of (X + 3Y) as follows:

$V(X+3Y) = V (X)+3^{2}V(Y)+2\times 3Cov (X,Y)\\=1+(9\times0.0625)+0\\=1.5625$

Then the standard deviation of (X + 3Y) is:

$SD(X + 3Y) =\sqrt{V(X + 3Y)}\\\sqrt{1.5625}\\=1.25$

Thus, the expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.

(b)

The amount of ice-cream left in the box after scooping out one scoop is represented as follows:

X - Y.

Compute the expected value of (X - Y) as follows:

$E(X-Y)=E(X)-E(Y)\\=48-2\\=46$

Compute the variance of (X - Y) as follows:

$V(X - Y) =V(X)+V(Y)-2Cov(X,Y)\\=1+0.0625-0\\=1.0625$

Then the standard deviation of (X - Y) is:

$SD(X-Y) =\sqrt{V(X -Y)}\\\sqrt{1.0625}\\=1.031$

Thus, the expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.

(c)

The variance of the sum or difference of two variables is computed by adding the individual variances. This is because the variance of each variable is dependent on the others.

7 0

2 years ago

A rectangular pool 18 meters by 12 meters is surrounded by a walkway of width x

vesna_86 [32]

Answer:

x=3 meters

Step-by-step explanation:

step 1

Find the area of the rectangular pool

$A=LW$

we have

$L=18\ m\\W=12\ m$

substitute

$A=18(12)=216\ m^2$

step 2

Find the area of rectangular pool including the area of the walkway

Let

x ----> the width of the walkway

we have

$L=(18+2x)\ m\\W=(12+2x)\ m$

substitute

$A=(18+2x)(12+2x)$

step 3

Find the area of the walkway

To find out the area of the walkway subtract the area of the pool from the area of rectangular pool including the area of the walkway

so

$A=(18+2x)(12+2x)-216$

step 4

Find the value of x if the area of the walkway equal the area of the pool

so

$(18+2x)(12+2x)-216=216$

Solve for x

$(18+2x)(12+2x)=432\\216+36x+24x+4x^{2}=432\\4x^{2} +60x-216=0$

Solve the quadratic equation by graphing

The solution is x=3 meters

see the attached figure

8 0

2 years ago

Louden County Wildlife Conservancy counts butterflies each year. Data over the last three years regarding four types of butterfl

MrRa [10]

Incomple question. However, here's the remaining part of the question:

14

2009

Meadow Fritillary= 5

Variegated Fritillary= 7

Zebra Swallowtail= 33

Eastern-Tailed Blue= 242

Louden County Butterfly Count

2010

Meadow Fritillary 34

Variegated Fritillary 95

Zebra Swallowtail 21

Eastern-Tailed Blue 168

2011

Meadow Fritillary

Variegated Fritillary

Zebra Swallowtail

Eastern-Tailed Blue

10

170

Options:

A) All butterfly populations are steadily decreasing.

B)All butterfly populations were larger than usual in 2010.

C)The Eastern-Tailed Blue butterfly is more common than the others.

D)The Meadow Fritillary is equally common as the Variegated Fritillary

Answer:

C

Step-by-step explanation:

Looking through the above count data by Louden County Wildlife Conservancy from 2009 to 2011 we notice the Eastern- Trailed Blue butterfly has a higher count, which implies that the Eastern-Tailed Blue butterfly is more common than the other butterflies.

Therefore, we could infer from the samples, that the Eastern-Tailed Blue butterfly is more common than others from the records of the past 3 three years.

5 0

1 year ago

Sequence 1.5,3.9,6.3,8.7

Nitella [24]

Lololollolollllollllolloolollllll IM TRYING TO GET POINTS. THX

7 0

2 years ago

Read 2 more answers

Consider the probability distribution of x, where x is the number of job applications completed by a college senior through the

Oksi-84 [34.3K]

The expected number out of 1000 selected college seniors that completed 1 job application through the career centers is given by 0.011 x 1000 = 11 which is close to 14.

The expected number out of 1000 selected college seniors that completed 2 job application through the career centers is given by 0.115 x 1000 = 115 which is far away from 15.

The expected number out of 1000 selected college seniors that completed 3 job application through the career centers is given by 0.123 x 1000 = 123 which is close to 130.

Therefore, the result that would be suprising is "15 seniors completed 2 job applications through the career center."

3 0

2 years ago