Dora has two savings account. The

Describe any domain restrictions that make sense for each of these scenarios: The price of a stock fluctuates several times thro

mrs_skeptik [129]

Answer:

a) Restriction: days of the year (a discrete time)

b) The time the ball is on the air: t = 0, until the balls touchs the ground (any real value from t = 0 to t when the ball hits the ground)

c) Any positive real value: x ≥ 0

Explanation:

The domain is the set of elements that can be inputs of the function.

1. The price of a stock fluctuates several times throughout the year as a function of time.

The input variable is the time through out the year.

If you assume that the fluctuaion is recorded once a day, and not every second, the logical assumption is that the time is a day of the year.

Thus, the restricion is the days of the year.

You could represent them in different forms, like DD/MM/YY, or even as a number from 1 to 365 if it covers just one year.

2. A ball is hit into the air by a baseball player, rises until it reaches its highest point, and then falls into left field.

The domain is the time the ball is on the air, which may be any real value from t = 0, until the balls touchs the ground.

It is not restricted to integer numbers.

3. The volume of a spherical balloon increases as the balloon is blown up and decreases as air is let out of it.

The inputs of the function are amounts of air, which may be measured as number of moles.

The number of moles can be any real value equal or greater than zero.

Thus, the domain is the positive real values x ≥ 0.

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Solve <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3x%20%2B%204%7D%7B2%7D%20%20%3D%209.5" id="TexFormula1" title=" \frac{3x

ELEN [110]

Answer: 5

Step by step:

(3x+4)/2 = 9.5
3x+4 = 9.5*2
3x+4 = 19
3x = 19 -4
3x = 15
x = 5

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A kite has vertices at (2, 4), (5, 4), (5, 1), and (0, –1).

JulsSmile [24]

The answer is 16.8 units

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translate the sentence into an equation and solve it. If 4 times a number is decreased by 5,the result is 19.find the number. th

eduard

4x-5=19
+5 +5
4x=24
x=6

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3 years ago

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A researcher collected data on the hours of TV watched per day from a sample of five people of different ages. Here are the resu

frozen [14]

Answer:

1. The least squares regression is y = -0.1015·x + 6.51

2. The independent variable is b) age

Please see attached table

Step-by-step explanation:

The least squares regression formula is given as follows;

$\dfrac{\sum_{i = 1}^{n}(x_{i} - \bar{x})\times \left (y_{i} - \bar{y} \right ) }{\sum_{i = 1}^{n}(x_{i} - \bar{x})^{2}}$

We have;

$\bar x$ = 24

$\bar y$ = 4

$\Sigma (x_i - \bar x) (y_i - \bar y)$ = -79

$\Sigma (x_i - \bar x)^2$ = 778

$\therefore \hat \beta =\dfrac{\sum_{i = 1}^{n}(x_{i} - \bar{x})\times \left (y_{i} - \bar{y} \right ) }{\sum_{i = 1}^{n}(x_{i} - \bar{x})^{2}} = \frac{-79}{778} = -0.1015$

The least squares regression is y = -0.1015·x + α

∴ α = y -0.1015·x = 6 - (-0.1015 × 5) = 6.51

The least squares regression is thus;

y = -0.1015·x + 6.51

2. The independent variable is the age b)

3. Steps to create an ANOVA table with α = 0.05

The overall mean = (43 + 30 + 22 + 20 + 5 + 1 + 6 + 4 + 3 + 6 )/10 = 14

There are 2 different treatment = $df_{treat} = 2 - 1 = 1$

There are 10 different treatment measurement = $df_{tot} = 10 - 1 = 9$

$df_{res} = 9 - 1 = 8$

$df_{treat} + df_{res} = df_{tot}$

The estimated effects are;

$\hat A_1 = 24 - 14 = 10$

$\hat A_2 = 4 - 14 = -10$

$SS_{treat} = 10^2 \times 5 + (-10)^2 \times 5 =1000$

$\sum_{i}\SS_{row}_i = \sum_{i}\sum_{j} (y_{ij} - \bar y)= [(1 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (3 - 4)^2 + (6 - 4)^2] = 18$

$\sum_{i} S S_{row}_i = \sum_{i}\sum_{j} (y_{ij} - \bar y) ^2= [(43 - 24)^2 + (30 - 24)^2 + (22 - 24)^2 + (20 - 24)^2 + (5 - 24)^2] = 778$

$S S_{res} = \sum_{i} S S_{row}_i = 778 + 18 = 796$

$SS_{tot}$ = (43 - 14)² + (30 - 14)² + (22 - 14)² + (20 - 14)² + (5 - 14)² + (1 - 14)1² + (6 - 4 )² + (3 - 14)² + (6 - 14)² = 1796

$MS_{treat} = \dfrac{SS_{treat} }{df_{treat} } = \dfrac{1000}{1} = 1000$

$MS_{res} = \dfrac{SS_{res} }{df_{res} } = \dfrac{796}{8} = 99.5$

F- value is given by the relation;

$F = \dfrac{MS_{treat} }{MS_{res} } = \frac{1000}{99.5} = 10.05$

We then look for the critical values at degrees of freedom 1 and 8 at α = 0.05 on the F-distribution tables 5.3177

Hence; $F = 10.05 > F_{1,8}^{Krit}(5\%) = 5.3177$ , we reject the null hypothesis.

7 0

3 years ago