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3 years ago

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Sally’s income has increased by 10% and she decides to change her consumption of macaroni and cheese from 10 boxes to 8 boxes. H

er income elasticity of demand is ____ and the good is a _____ type of good.

1 answer:

UNO [17]3 years ago

6 0

Answer:

-2; Inferior good

Step-by-step explanation:

Given that,

Initial Quantity = 10 boxes

New Quantity = 8 boxes

Percentage increase in Sally's income = 10%

Change in consumption:

= 8 boxes - 10 boxes

= - 2 boxes

Percentage change in quantity demanded:

= (Change in quantity demanded ÷ Initial quantity) × 100

= (-2 ÷ 10) × 100

= - 20%

Therefore,

Income elasticity of demand:

= percentage change in quantity demanded ÷ Percentage change in income

= - 20% ÷ 10

= -2

Inferior goods are generally have a negative income elasticity of demand which means that an increase in the income of the consumer will lead to reduce the quantity demanded for inferior good and vice versa.

Hence, the good is a inferior type of good.

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2.) Fill in the blank.

Montano1993 [528]

Answer:

E.) 1

Step-by-step explanation:

Firstly we will solve for L.H.S.

L.H.S. = $Csc^2\theta$

Since we know that $Csc^2\theta$ is the inverse of $Sin^2\theta$ .

So we can say that;

$csc^2\theta=\frac{1}{sin^2\theta}$

Now For R.H.S.

$Cot^2\theta+1$

Since we can rewrite $cot^2\theta$ as $\frac{cos^2\theta}{sin^2\theta}$ .

Now we can say that the R.H.S. is;

$\frac{cos^2\theta}{sin^2\theta}+1$

Now we add the fraction and get;

$\frac{cos^2\theta+sin^2\theta}{sin^2\theta}$

Now according to trigonometric identity;

$cos^2\theta+sin^2\theta=1$

So, $\frac{cos^2\theta+sin^2\theta}{sin^2\theta}=\frac{1}{sin^2\theta}$

Here,

$csc^2\theta=\frac{1}{sin^2\theta}$ and $Cot^2\theta+1$ = $\frac{1}{sin^2\theta}$ <em> </em>

L.H.S. = R.H.S.

Hence $csc^2\theta=cot^2\theta+1$

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3 years ago

Calculate the area of a circle with a diameter of 12 cm

rosijanka [135]

we know its diameter is 12, thus its radius must be half that, or 6.

$\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=6 \end{cases}\implies A=\pi 6^2\implies A=36\pi \implies A\approx 113.097$

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What is the initial value of the exponential function represented by the table

rodikova [14]

Correct answer should be 1/2

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Katarina [22]

The answer will be “b” because exponent means mutiply

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Read 2 more answers

Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail inste

Vladimir [108]

Answer:

The value of t test statistics is 5.9028.

We conclude that the true mean is greater than 10 at the .01 level of significance.

Step-by-step explanation:

We are given that a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail instead.

The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages.

<em />

<em>Let </em> $\mu$ <em> = true mean transmission of pages.</em>

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 10 pages {means that the true mean is smaller than or equal to 10}

Alternate Hypothesis, $H_A$ : $\mu$ > 10 pages {means that the true mean is greater than 10}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 14.44 pages

s = sample standard deviation = 4.45 pages

n = sample of fax transmissions = 35

So, <u><em>test statistics</em></u> = $\frac{14.44-10}{\frac{4.45}{\sqrt{35} } }$ ~ $t_3_4$

= 5.9028

(a) The value of t test statistics is 5.9028.

Now, at 0.01 significance level the z table gives critical values of 2.441 at 34 degree of freedom for right-tailed test.

<em>Since our test statistics is more than the critical values of z as 5.9028 > 2.441, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis.</u></em>

Therefore, we conclude that the true mean is greater than 10.

(b) Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_3_4$ > 5.9028) = Less than 0.05% {using t table)

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