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Nonamiya [84]
2 years ago
13

There is a harvest festival each year on the planet Bozone. During that time, most of the Bozone residents sit down to rejoice a

nd eat roast snig. Past history shows that a reasonable price-demand equation is: p = 300 − 18 x where x is the number of kilograms of snig sold, and p is the price per kilogram of snig in the local currency, the Boat. a . To sell an additional kilogram of snig, how much does the price need to decrease
Mathematics
2 answers:
Oliga [24]2 years ago
8 0

Answer:

The price would be decreased by 18 bozats

Step-by-step explanation:

The following information is given in the question

x = number of kilograms of snig sold

P = Price per kilogram

And, the equation is

p = 300 - 18x

Now if an extra kilogram is sold so it should be x+1

Now the new price is

New price = 300 - 18(x + 1)

= 300 - 18x - 18

Therefore the price would be decreased by 18 bozats

Lina20 [59]2 years ago
7 0

Given that:

Price-demand equation is

p=300-18x

where, x is the number of kilograms of snig sold, and p is the price per kilogram of snig in the local currency, the Boat.

To find:

The price which needs to decrease to sell an additional kilogram of snig,

Solution:

We have,

p=300-18x

If x=1, then

p=300-18(1)

p=300-18

p=282

It means, the price is 282 if 1 kg of snig sold.

Similarly,

If x=2, then

p=300-18(2)

p=300-36

p=264

It means, the price is 264 if 2 kg of snig sold.

Change in price if sales increased by 1 kg is

264-282=-18

Here, negative sign means decrease in price.

Therefore, the price needs to decrease by 18 to sell an additional kilogram of snig.

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A statistics practitioner in a large university is investigating the factors that affect salary of professors. He wondered if ev
Ulleksa [173]

Answer:

Step-by-step explanation:

Hello!

Given the linear regression of Y: "Annual salary" as a function of X: "Mean score on teaching evaluation" of a population of university professors. It is desired to study whether student evaluations are related to salaries.

The population equation line is

E(Y)= β₀ + β₁X

Using the information of a n= 100 sample, the following data was calculated:

R²= 0.23

                Coefficient    Standard Error  

Intercept    25675.5           11393

x                  5321                  2119

The estimated equation is

^Y= 25675.5 + 5321X

Now if the interest is to test if the teaching evaluation affects the proffesor's annual salary, the hypotheses are:

H₀: β = 0

H₁: β ≠ 0

There are two statistic you can use to make this test, a Student's t or an ANOVA F.

Since you have information about the estimation of  β you can calculate the two tailed t test using the formula:

t= \frac{b - \beta }{\frac{Sb}{\sqrt{n} } } ~t_{n-2}

t= \frac{ 5321 - 0 }{\frac{2119}{\sqrt{100} } } = 25.1109

The p-value is two-tailed, and is the probability of getting a value as extreme as the calculated t_{H_0} under the distribution t_{98}

p-value < 0.00001

I hope it helps!

3 0
2 years ago
A circle has diameter of 11cm
dlinn [17]

Since length of diagonal (  Diagonal= 9.9cm ) is less than diameter of circle ( 11 cm ) , Therefore , the square will fit inside the circle without touching the edge of the circle.

<u>Step-by-step explanation:</u>

Here we have , A circle has diameter of 11 cm  A square has side length of 7 cm . Use Pythagoras’ Theorem to show that the square will fit inside the circle without touching the edge of the circle . Let's find out:

We know the concept that for any square to fit inside the circle without touching the edge of circle  , diagonal of square must be less than diameter of circle  . Let's find out length of diagonal by using Pythagoras Theorem :

Hypotenuse ^2 = Perpendicular^2+Base^2

For a square , Perpendicular = base = side

⇒ Diagonal^2 = 2(side)^2

⇒ Diagonal= \sqrt{2}(side)

⇒ Diagonal= \sqrt{2}(7)

⇒ Diagonal= 9.9cm

Since length of diagonal (  Diagonal= 9.9cm ) is less than diameter of circle ( 11 cm ) , Therefore , the square will fit inside the circle without ruching the edge of the circle.

5 0
2 years ago
an ellipse has a center at the origin, a vertex along the major axis at (13, 0), and a focus at (12, 0). What is the equation of
yan [13]
Hi, as the elipce has a center at the origion and tha major distance is the axis

\frac{x^2}{a^2} + \frac{y^2}{b^2}  = 1

But, 

c = 12        Distance of focus

a = 13   ->  Major distance

b = ?  -> 

But, 

\\ b^2 + c^2 = a^2&#10; \\ &#10; \\ b^2 + 12^2 = 13^2&#10; \\ &#10; \\ b^2 + 144 = 169&#10; \\ &#10; \\ b^2 = 169 - 144&#10; \\ &#10; \\ b^2 = 25&#10; \\ &#10; \\ b =  \sqrt{5} &#10; \\ &#10; \\ b = 5&#10;&#10;

Then,


\\  \frac{x^2}{(13)^2} +  \frac{y^2}{(5)^2} = 1&#10; \\ &#10; \\ or&#10; \\ &#10; \\  \frac{x^2}{169} +  \frac{y^2}{25} = 1
5 0
2 years ago
Read 2 more answers
A circle has a circumference of \blue{12}12start color #6495ed, 12, end color #6495ed. It has an arc of length \dfrac{8}{5} 5 8
Thepotemich [5.8K]

Answer:

Therefore,

The Central Angle is 48°.

Step-by-step explanation:

Given:

Circumference = 12 units

Arc\ length = \dfrac{8}{5}= 1.6\ unit

To Find:

Central angle = θ = ?

Solution:

If "θ" the central angle is in degree then arc of length is given by

Arc\ length = \dfrac{\theta}{360}\times Circumference

Substituting the values we get

1.6 = \dfrac{\theta}{360}\times 12\\\\\theta = 48\°

Therefore,

The Central Angle is 48°.

6 0
2 years ago
Read 2 more answers
On this graph, 4:00 p.m. occurs at 16 hours after midnight, and 6:00 p.m. occurs at 18 hours after midnight. Which statements ar
KatRina [158]

Answer:

TRUE OPTIONS ARE:

<em>"The temperature increased until 4:00 p.m. "</em>

<em>"The temperature decreased after 6:00 p.m. "</em>

<em>"The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m."</em>

Step-by-step explanation:

<em>graph attached and complete question below:</em>

<em>Which statements are true about the temperatures Luis  recorded on the graph? Check all that apply. </em>

  • <em>The temperature increased until 4:00 p.m. </em>
  • <em>The temperature was not recorded between 4:00 p.m.  and 6:00 p.m. </em>
  • <em>The temperature decreased after 6:00 p.m. </em>
  • <em>The temperature increased and then decreased before  holding constant. </em>
  • <em>The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m.</em>

<em />

Until 1600 hours, the graph increases, so we can say temperature increased until 4.00 pm (FIRST OPTION TRUE).

From 1600 to 1800 hours (4 - 6pm), the temperature stayed same (horizontal line). This doesn't mean the temperature wasn't recorded. (2nd OPTION FALSE).

After 1800 hours (6pm), the line goes downward, so temperature decreased after 6 pm. (3rd OPTION TRUE).

If you look at the temperature graph, we can see temperature increased, then increased more, then constant, then decreased. Thus the 4th option isnt true. (4th OPTION FALSE).

Before 12, the increase isn't as sharp as after 12. After 12 temperature increase has more slope, so this increase is more. (5th OPTION TRUE).

7 0
2 years ago
Read 2 more answers
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