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BigorU [14]
3 years ago
7

Many communities today are concerned with their local government’s ability to appropriately manage the taxes it collects. If a g

overnment spends more money on expenditures than it collects in taxes, it may end up with not enough money, called deficit. Which of the following could result in a county government budget in deficit?
a.The government raises sales and property taxes without adjusting expenditures.

b.The county closes a fire station, reducing expenditures, without adjusting taxes.

c.The county gives all public school teachers a $2,000 raise without adjusting taxes.

d.The county collects donations to build a fountain in city hall.
Mathematics
2 answers:
sleet_krkn [62]3 years ago
8 0
D.  the county collects donations to build a fountain in city hall.

zysi [14]3 years ago
3 0

Answer:

the answer is D

Step-by-step explanation:

i just took the test

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An environmental agency worries that many cars may be violating clean air emissions standards. The agency hopes to check a sampl
goldenfox [79]

Answer:

197

Step-by-step explanation:

Sample proportion is; p^ = 12/50 = 0.24

Margin of error of 5% is given by the formula;

ME = (z_0.05) √[p^(1 - p^)/n]

Let's make n(number of sample) the subject.

n = ((z_0.05)/ME)²((p^)(1 - p^))

From tables, the z-score of 0.05 is 1.645

Thus;

n = ((1.645)/0.05)²(0.24(1 - 0.24))

n ≈ 197

3 0
3 years ago
which of the following is a binomial experiment? a) flipping a coin until it comes up tails b) flipping a coin until 8 heads are
Luda [366]
A binomial experiment must have all of the following:
a) A fixed number of trials.
b) The result of each trial is not affected by the results of the other trials (independence).
c) The probability of a 'success' is the same for each trial.
d) Each trial has only two possible outcomes, which are usually called 'success' and 'failure'.
Using the above criteria, the only choice of answer that qualifies as a binomial experiment is c) "flipping a coin 10 times and recording if it comes up heads."
4 0
3 years ago
Read 2 more answers
Técnica de conteo en estadística 6. Regla multiplicativa. Para el experimento de un lanzamiento de dado y sacar una carta de una
babymother [125]

Answer:

Para el punto a. tenemos 312 elementos en el espacio muestral

Para el punto b. la probabilidad pedida es \frac{1}{26}

Step-by-step explanation:

Comencemos explicando el principio de multiplicación. Supongamos que se tiene un experimento por etapas, es decir, una primera etapa que se puede llevar a cabo de '' x_{1} '' maneras distintas, una segunda etapa que se puede llevar a cabo de '' x_{2} '' maneras distintas y así sucesivamente hasta una etapa final que se puede llevar a cabo de '' x_{n} '' maneras distintas. La cantidad total de formas de desarrollar todo el experimento se puede calcular como el producto de las distintas maneras en que se puede ejecutar cada etapa :

Formas de desarrollar todo el experimento = x_{1}x_{2} ... x_{n}

Para el punto a. del ejercicio vamos a utilizar este principio para calcular la cantidad de elementos del espacio muestral. La primera etapa del experimento sería el lanzamiento del dado y una segunda etapa el sacar una carta de una baraja de poker (52 cartas distintas). Entonces, para un dado tenemos 6 posibles opciones y cuando sacamos una carta podemos obtener 52 resultados distintos. Por lo tanto ⇒

(6)(52)=312

Tenemos un total de 312 elementos en el espacio muestral (todos distintos) y con la misma probabilidad de ocurrencia ya que suponemos que en el dado tenemos la misma probabilidad de obtener un uno, un dos, etc; así como en la baraja tenemos la misma probabilidad de sacar cada una de las cartas.

Para el punto b. utilizaremos el hecho de que estamos frente a un espacio muestral equiprobable y calcularemos la probabilidad del evento ''sale un número par en el dado y un AS en la baraja'' contando los casos favorables a este evento y dividiendo por la cantidad de elementos totales del espacio muestral. Podremos hacerlo de esta manera ya que el espacio muestral es equiprobable. Definimos para ello

A: ''Sale un número par en el dado y un AS en la baraja''

P(A)=\frac{CasosFavorablesAlEventoA}{CasosTotales}

Los casos totales son 312 (calculados en el inciso a.)

Los casos favorables al evento A se pueden escribir utilizando el principio de multiplicación como

(3)(4)=12

El 3 es por los tres números pares que podemos obtener del dado (estos números son el dos, cuatro y el seis).

El 4 se debe a la cantidad de ases en una baraja de Póker de 52 cartas (estos son el as de corazones, el as de picas, el as de trébol y el as de diamantes).

Reemplazando en la expresión de P(A) obtenemos :

P(A)=\frac{12}{312}=\frac{1}{26}

Otra forma de haber calculado esta probabilidad es utilizando el hecho de que tirar el dado y sacar una carta de la baraja son dos eventos independientes. Por ende, la probabilidad se puede calcular como el producto de la probabilidad parcial de cada evento. Esto es :

B : ''Sale un número par en el dado''

C : ''Saco un as en la baraja''

P(B)=\frac{3}{6}=\frac{1}{2}

Tres casos favorables a que salga un número par de seis posibles.

P(C)=\frac{4}{52}

Cuatro casos favorables a obtener un as de la baraja de 52 casos posibles.

Ahora escribimos

P ( B ∩ C ) = P(B).P(C)

Por ser B y C dos eventos independientes. Entonces

P(B)P(C)=(\frac{1}{2})(\frac{4}{52})=\frac{1}{26}

Obteniéndose la misma probabilidad ya calculada.

8 0
3 years ago
A child in a neighborhood had a lemonade stand where large cups sold for $0.75, medium cups sold for $0.60 and small cups for $0
chubhunter [2.5K]

Answer:

  • s + m + L = 78
  • 0.45s +0.60m +0.75L = 46.95
  • s - m + L = 4

Step-by-step explanation:

The problem statement asks for the number of each size cup sold. You are told there are three sizes, so it is convenient to use one variable to represent the number of cups of a given size that were sold—three variables total.The problem statements tell you the relationships between the numbers:

  • the total number of cups sold
  • the total value of cups sold
  • the relationship between numbers of one size versus other sizes

It is appropriate to write an equation for each of these relationships, for a total of three equations. (Here, the last of these equations is put into standard form to make it easier to translate this to a matrix equation for solving by calculator.)

___

The solution is: 20 small, 37 medium, and 21 large.

4 0
3 years ago
Mr.Williams' physical education class lasts 7/8 hour.How many minutes are not spent on instructions . playing game:1/2 , instruc
alexandr1967 [171]
First you would find a common denominator. If 40 is the LCM then you would have to add playing games, instruction,warm-up, and cool-down. Then you will subtract the sum from the total class time and bam! You get your answer!
7 0
3 years ago
Read 2 more answers
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