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2 years ago

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Trevor purchases a car. The value of the car is modeled by the function y=22000(0.86)t. Which statement below best describes the

value the base 0.86?
Group of answer choices

The car Trevor purchased depreciates at a rate of 14%.

The car that Trevor purchased appreciates at a rate of 14%.

The car that Trevor purchased appreciates at a rate of 86%.

The car Trevor purchased depreciates at a rate of 86%.

1 answer:

stealth61 [152]2 years ago

5 0

The statement below best describes the value the base 0.86 is that "the car Trevor purchased depreciates at a rate of 86%".

It should be noted thàt depreciation occurs when we buy an equipment or machine and use it. With time, due to wear and tear or usage, the value reduces.

The value of the car will not appreciate but rather depreciate. Therefore, the correct option is "the car Trevor purchased depreciates at a rate of 86%".

In conclusion, the correct option is D.

Read related link on:

brainly.com/question/21247090

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The Florida Everglades welcomes about 2x10 to the third power visitors per day. Based on this, about how many visitors come to t

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2,000 is the correct answer I am sure of it

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2 years ago

The Anza-Borrego Desert State Park is one of the best places in the United States for viewing stars. During one 50-day period, c

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22

Step-by-step explanation:

I would assume that the 6% rate would carry for the rest of the year, i.e. all 365 days, 6% of those have clouds interfering with stargazing

365 * 6% = 365 * 0.06 = 21.9, and round up to 22 because you can't really have .9 of a day here.

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2 years ago

A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The t

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Answer:

Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

$\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5$

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

$\widehat{Y}=Y_1+Y_2+...+Y_{200}$

= $200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs$

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

S= $\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486$

Then, the standard error of $\widehat{Y} is$

$\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:$ =83.785

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]$

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

$\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45$

The sample ratio is

$r=\frac{\overline{y}}{\overline{x}}=\frac{4.5}{45}$ =0.1 , this is also the estimate of the population ratio R= $\frac{\overline{Y}}{\overline{X}} .$

Therefore, the estimated total weight of grain in the population using ratio estimator is

$\widehat{Y}_R\: \: =r\times 8800\: \: =0.1\times 8800\: \: =880\,$ lbs

4) The variance of the ratio estimator is

var(r)= $\frac{N-n}{N}\frac{1}{n}\frac{1}{\mu_x^2}\frac{\sum_{i=1}^{5}(y_i-rx_i)^2}{n-1} , where \mu_x=8800/200=44lbs$

= $\frac{200-5}{200}\, \frac{1}{5}\: \frac{1}{44^2}\, \frac{0.2}{5-1}=0.000005$

Hence, the standard error of the estimate of the total population is

$\sigma_R=\sqrt{X^2 \: var(r)}\: \: \: =\sqrt{8800^2\times 0.000005}\: \: \:$ =21.556

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]$

8 0

2 years ago

Read 2 more answers

The number of tickets purchased by an individual for Beckham College's holiday music festival is a uniformly distributed random

egoroff_w [7]

Answer:

The mean is 4.5 and the standard deviation is 1.44.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean of the uniform probability distribution is:

$M = \frac{(a + b)}{2}$

The standard deviation of the uniform probability distribution is:

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

Uniformly distributed random variable ranging from 2 to 7.

This means that $a = 2, b = 7$ .

So

$M = \frac{(2 + 7)}{2} = 4.5$

$S = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(7 - 2)^{2}}{12}} = 1.44$

The mean is 4.5 and the standard deviation is 1.44.

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2 years ago

Imaginá que tenés 125 dados cúbicos del mismo tamaño ¿Cuantos dados de altura tiene el cubo de mayor tamaño que podés armar apil

kumpel [21]

Answer:

(i) Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

Step-by-step explanation:

(i) Sabemos por la Geometría Euclídea del Espacio que un cubo es un sólido regular con 6 caras cuadradas y longitudes iguales. Cada dado tiene un volumen de 1 dado cúbico y 125 dados dan un volumen total de 125 dados cúbicos.

El volumen de un cubo está dado por la siguiente fórmula:

$V = L^{3}$

Donde:

$L$ - Longitud de la arista, medida en dados.

$V$ - Volumen del cubo, medido en dados cúbicos.

Ahora, necesitamos despejar la longitud de la arista para calcular la altura máxima posible:

$L = \sqrt[3]{V}$

Dado que $V = 125\,dados^{3}$ , encontramos que la altura del cubo de mayor tamaño sería:

$L =\sqrt[3]{125\,dados^{3}}$

$L = 5\,dados$

Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) El área cuadrada formada por cubos está determinada por la siguiente fórmula:

$A = L^{2}$

Donde:

$L$ - Longitud de arista, medida en dados.

$A$ - Área, medida en dados cuadrados.

Puesto que la longitud de arista se basa en un conjunto discreto, esto es, el número de dados disponibles, debemos encontrar el valor máximo de $L$ tal que no supere 125 y de un área entera. Es decir:

$L \leq 125\,dados$

Si cada cubo tiene un área de 1 dado cuadrado, entonces un cuadrado conformado por 125 dados tiene un área total de 125 dados cuadrados. Entonces:

$L^{2}< 125\,dados^{2}$

Esto nos lleva a decir que:

$L < 11.180\,dados$

Entonces, la longitud máxima del cuadrado con la mayor cantidad de cubos posible es de 11 dados. El número total requerido de cubos es el cuadrado de esa cifra, es decir:

$n = (11\,dados)^{2}$

$n = 121\,dados$

Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

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2 years ago