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Anon25 [30]
3 years ago
15

Choose all that are sources of funding for social services in a society. income tax collection service fees collection traffic t

ickets port fees
Mathematics
2 answers:
AfilCa [17]3 years ago
8 0

Answer:  Second option is correct.

Step-by-step explanation:

Sources of funding for social services in a society are as follows:

Service fees collection is the charges for service that are charged by society for social services.

All other fees are taken by government instead of society.

Income tax collection is the direct tax levied by the government on a particular person.

Traffic tickets are imposed by the government as non - revenue receipt of government.

Port fees are also imposed by the government as non revenue receipt of government.

Hence, Second option is correct.

oksian1 [2.3K]3 years ago
5 0

Social services can be defined as the services which are provided by the government for the welfare and benefit of the community, such as education, medical care, and housing.

So, the sources of funding are : income tax collection and service fees collection.

You might be interested in
In the triangle above, the measure of A is 66°, and the measure of B is 68°. What is the measure of C?
valentina_108 [34]

Answer:

46°

Step-by-step explanation:

Given that :

Angle, A = 66°

Angle B = 68°

The sum of angles in a triangle is given by :

A + B + C = 180° (sum if angles una triangle)

66° + 68° + C = 180°

134° + C = 180°

C = 180° - 134°

C = 46°

Hence, angle C is 46°

6 0
3 years ago
The Empirical Rule The following data represent the length of eruption for a random sample of eruptions at the Old Faithful geys
ad-work [718]

Answer:

(a) Sample Standard Deviation approximately to the nearest whole number = 6

(b) The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is.

(c) The percentage of eruptions that last between 92 and 116 seconds using the empirical rule is 95%

(d) The actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) The percentage of eruptions that last less than 98 seconds using the empirical rule is 16%

(f) The actual percentage of eruptions that last less than 98 seconds is 15.866%

Step-by-step explanation:

(a) Determine the sample standard deviation length of eruption.

Express your answer rounded to the nearest whole number.

Step 1

We find the Mean.

Mean = Sum of Terms/Number of Terms

= 90+ 90+ 92+94+ 95+99+99+100+100, 101+ 101+ 101+101+ 102+102+ 102+103+103+ 103+103+103+ 104+ 104+104+105+105+105+ 106+106+107+108+108+108 + 109+ 109+ 110+ 110+110+110+ 110+ 111+ 113+ 116+120/44

= 4582/44

= 104.1363636

Step 2

Sample Standard deviation = √(x - Mean)²/n - 1

=√( 90 - 104.1363636)²+ (90-104.1363636)² + (92 -104.1363636)² ..........)/44 - 1

= √(199.836777 + 199.836777 + 147.2913224+ 102.7458678+ 83.47314049+ 26.3822314+ 26.3822314+ 17.10950413+17.10950413+ 9.836776857+ 9.836776857, 9.836776857+9.836776857+ 4.564049585+ 4.564049585+ 4.564049585+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 0.01859504133+ 0.01859504133+ 0.01859504133+ 0.7458677685+ 0.7458677685+ 0.7458677685+ 3.473140497+ 3.473140497+ 8.200413225+ 14.92768595+ 14.92768595+ 14.92768595+ 23.65495868+ 23.65495868+ 34.38223141+ 34.38223141+34.38223141+ 34.38223141+ 34.38223141+47.10950414+ 78.56404959+ 140.7458677+ 251.6549586) /43

= √1679.181818/43

= √39.05073996

= 6.249059126

Approximately to the nearest whole number:

Mean = 104

Standard deviation = 6

(b) On the basis of the histogram drawn in Section 3.1, Problem 28, comment on the appropriateness of using the Empirical Rule to make any general statements about the length of eruptions.

The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is .

(c) Use the Empirical Rule to determine the percentage of eruptions that last between 92 and 116 seconds.

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

Mean = 104, Standard deviation = 6

For 68% μ - σ = 104 - 6 = 98, μ + σ = 104 + 6 = 110

For 95% μ – 2σ = 104 -2(6) = 104 - 12 = 92

μ + 2σ = 104 +2(6) = 104 + 12 = 116

Therefore, the percentage of eruptions that last between 92 and 116 seconds is 95%

(d) Determine the actual percentage of eruptions that last between 92 and 116 seconds, inclusive.

We solve for this using z score formula

The formula for calculating a z-score is is z = (x-μ)/σ

where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean = 104, Standard deviation = 6

For x = 92

z = 92 - 104/6

= -2

Probability value from Z-Table:

P(x = 92) = P(z = -2) = 0.02275

For x = 116

z = 92 - 116/6

= 2

Probability value from Z-Table:

P(x = 116) = P(z = 2) = 0.97725

The actual percentage of eruptions that last between 92 and 116 seconds

= P(x = 116) - P(x = 92)

= 0.97725 - 0.02275

= 0.9545

Converting to percentage = 0.9545 × 100

= 95.45%

Therefore, the actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) Use the Empirical Rule to determine the percentage of eruptions that last less than 98 seconds

The empirical rule formula:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

For 68% μ - σ = 104 - 6 = 98,

Therefore, 68% of eruptions that last for 98 seconds.

For less than 98 seconds which is the Left hand side of the distribution, it is calculated as

= 100 - 68/2

= 32/2

= 16%

Therefore, the percentage of eruptions that last less than 98 seconds is 16%

(f) Determine the actual percentage of eruptions that last less than 98 seconds.

The formula for calculating a z-score is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

For x = 98

Z score = x - μ/σ

= 98 - 104/6

= -1

Probability value from Z-Table:

P(x ≤ 98) = P(x < 98) = 0.15866

Converting to percentage =

0.15866 × 100

= 15.866%

Therefore, the actual percentage of eruptions that last less than 98 seconds is 15.866%

4 0
3 years ago
Find the solution to the equation or inequality below. You must type in the entire solution, for example x=5 if it is an equatio
Alina [70]
5x - 9 = 26
5x - 9 + 9 = 26 + 9
5x = 35
5x/5 = 35/5
X = 7
4 0
3 years ago
What is the relationship between the lines determined by the following two equations?
oksano4ka [1.4K]

Answer:

C. They are the same line.

Step-by-step explanation:

In order to compare the linear equations given, they need to be in the same form.  The best form in order to evaluate slope and y-intercept is slope-intercept form, y = mx + b.  Since the second equation is already in slope-intercept form, we need to use inverse operations to convert the first equation:

6x - 2y = 16 ----  6x - 2y - 6x = 16 - 6x ----  -2y = -6x + 16

-2y/-2 = -6x/-2 + 16/-2

y = 3x - 8

Since both equations are in the form y = 3x - 8, then they are both the same line.

4 0
3 years ago
*15 points help easy question*
DIA [1.3K]

Answer:

x = 12

Step-by-step explanation:

The segment from the vertex to the base is a perpendicular bisector.

Using Pythagoras' identity on the right triangle on the right.

(\frac{1}{2} x )² + 3² = (\sqrt{45} )² , that is

\frac{1}{4} x² + 9 = 45 ( subtract 9 from both sides )

\frac{1}{4} x² = 36 ( multiply both sides by 4 )

x² = 144 ( take the square root of both sides )

x = \sqrt{144} = 12

8 0
3 years ago
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