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1 year ago

Which statement best compares the variability of the number of snacks grabbed for Class A and Class B?

Mathematics

1 answer:

UNO [17]1 year ago

5 0

*see attachment below for the dot plots

Answer:

A. The number of snacks grabbed for Class A has less variability than the number of snacks grabbed for Class B.

Step-by-step explanation:

The extent of variability of the data represented on a dot plot can be determined by the range value.

Range is the difference between maximum value of the data set and the minimum value of the data set.

The larger the range, the more the data has more variability, while the smaller the range, the less the variability of the data.

Range for Class A = 37 - 15 = 22

Range for Class B = 45 - 12 = 33

Therefore, class B has more variability in the number of snacks grabbed compared to Class A.

Correct statement is: "The number of snacks grabbed for Class A has less variability than the number of snacks grabbed for Class B."

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A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

2 years ago

Rylee took out a loan for $3600 at 13% interest, compounded annually. If she makes yearly payments of $460, will she ever pay of

elena55 [62]

Given that Rylee took out a loan for $3600 at 13% interest.

Where interest is compounded annually.

Interest for 1 year = 13% of 3600 = 0.13*3600 = 468

Amount due after 1 year = Loan + interest = 3600+468 = 4068

Monthly payment = 460

So Amount to be paid after 1 year = 4068-460 = 3608

New due amount $3608 is more than the loan amount $3600

Which means loan will always remain due for his entire life.

Hence Rylee will never be able to pay off the loan.

Interest must be less than the monthly payment in order to pay off the loan.

3 0

2 years ago

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I need to find both of the solutions to the equation 100+(n-2)^ = 149

rodikova [14]

Answer:

Step-by-step explanation:

hello :

100+(n-2)² = 149

100-100+(n-2)² = 149-100

(n-2)² = 49

(n-2)² - 49 =0 but 49=7²

(n-2)² - 7² =0 use identity : a²-b²=(a-b)(a+b)

(n-2-7)(n-2+7)=0

(n-9)(n+5)=0

n-9=0 or n+5=0

n=9 or n=-5

5 0

2 years ago

There is a 0.9991 probability that a randomly selected 31-year-old male lives through the year. A life insurance company charge

jarptica [38.1K]

Answer:

a) Monetary values corresponding to the two events are:

-In case of surviving the year = -166$

-In case of a death in the year = 89834$

b) Expected value of the purchasing the insurance is -85 $

c) Yes, insurance company can make a profit with this policy.

Step-by-step explanation:

a) The man need to pay 166$ first to enroll the insurance policy. If he survives within a year, he will lose 166$. Otherwise, if he dies within a year he will profit 89834$.

b) Expected value of the purchasing the insurance as following:

-In case of surviving the year:

Value: -166$

Probability: 0,9991

-In case of death in a year

Value: 89834$

Probability: 0,0009

Expected value is E(x) = -166×0,9991 + 89834×0,0009 = -85 $

c) Lets consider that 10000 different 31 year old man enrolled to this insurance policy. According to probability of death, 9 out of 10000 man expected to be dead within the year. Therefore, company need to pay 9*90000 = 810000$ to their costumers. But, company will collect 10000*166=1660000$ from their costumers in the beginning of the year

So, it is expected that company is going to profit 1660000-810000=850000$ per year.

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2 years ago

The figure is made up of a cylinder and a half sphere. What is the volume of the composite figure? Use 3.14 for Pi. Round to the

vodomira [7]

Answer:

Step-by-step explanation:

6 0

2 years ago

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