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3 years ago

11

Stanley drove his car on a business trip when he left, the mileage was 840 miles, and when he returned, the mileage was 1,200 mi

les. The car used 12 gallons of gasoline for this trip.
Carla made the same trip as Stanley, but her car used only 10 gallons of gasoline. How do the slopes for Stanley's and Carla's cars compare?

1 answer:

Yuri [45]3 years ago

5 0

Stanley's unit rate is 30 miles per gallon ( represented by: $y=30x$ )
Carla's unit rate is 36 miles per gallon ( represented by: $y=36x$ )

Carla's unit rate is higher than Stanley's. (mention the slope being steeper on the graph, because I have the same vacation homework.)

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On Sasha's desk there are c green notebooks. The number of red notebooks on her desk is 6 less than number of green notebooks. H

aksik [14]

Sasha's desk has $c$ green notebooks.

The number of red notebooks on her desk is $c-6$ .

Sasha's desk has either green or red notebooks.

The total number of notebooks on Sasha's notebook is $c+c-6=2c-6$ .

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2 years ago

Exhibit 11-2 We are interested in determining whether the variances of the sales at two music stores (A and B) are equal. A samp

Fiesta28 [93]

Answer:

Option A , should not be rejected

Step-by-step explanation:

From the question we are told that:

Sample 1 $n=25$

Standard deviation 1 $\sigma_1=30$

Sample 2 $n=16$

Standard deviation 2 $\sigma_2=20$

Level of confidence $\alpha=95\%=>0.95$

Generally The Hypothesis are given as

$H_0:\sigma_1^2=\sigma_1^2$

$H_1:\sigma_1^2\neq \sigma_1^2$

Generally the equation for test Statistics is mathematically given by

$T=\frac{\sigma_1^2}{\sigma_2^2}$

$T=\frac{30^2}{20^2}$

$T=2.25$

Therefore

Critical Value

$X=(\alpha,df_1)$

$X'=(\alpha,df_2)$

Where

$df=n-1$

Therefore

$X=(0.95,24)$

$X'=(0.95,15)$

From Table

$T_{Critical}=T_x,T_{x'}$

$T_{Critical}=0.41,2.701$

Therefore

$T_x$

Hence,We fail to reject the Null Hypothesis $H_0$

Option A , should not be rejected

3 0

1 year ago

Consider the first five steps of the derivation of The Quadratic function

Burka [1]

Answer:

Step-by-step explanation:

pick like terms

x² + b²/4a² = -c / a + b²/4a²

b²/4a² = (b/2a)²

x² + (b/2a)² = -c/a + (b/2a)²

(x + b/2a)² = -c/a + (b/2a)² = -c / a + b²/4a² = (-4ac+ b²)/4a²

(x + b/2a)² = (-4ac+ b²)/4a²

square root both sides

√{(x + b/2a)²} = √{(-4ac+ b²)/4a²}

x + b/2a = √(-4ac+ b²) / √(4a²) = √(-4ac+ b²) / 2a = √( b²-4ac) / 2a

x + b/2a = √( b²-4ac) / 2a

subtract b/2a from both sides

x + b/2a -b/2a = {√( b²-4ac) / 2a } -b/2a

x = -b/2a + {√( b²-4ac) / 2a }

the l.c.m is the same

x = {-b±√( b²-4ac)}/2a

7 0

2 years ago

Nora is running a race that is 26.2 miles. She is running at a speed of 8 miles per hour. She has completed 3/4 of the race. How

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The answer to this problem would be about 4 hours

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2 years ago

A bicycle wheel with diameter 16 inches rides over a screw in the street. The screw is on level ground before it punctures the b

Julli [10]

Answer:

The correct option is;

16 inches

Step-by-step explanation:

The parameters of the motion given are;

The diameter, D of the bicycle = 16 inches;

The distance the bike moves (forward) after the screw punctures the tire = 56·π inches

We note that the circumference of the bicycle = π·D = π × 16 = 16·π inches

Therefore;

56·π inches/(16·π inches) = 3.5

Showing that the bicycle moves three and half complete turns (revolution) where after each complete turn, the screw starts from the bottom of the tire.

The height, h of the screw in the final half turn is given by the relation;

h = A×cos(Bx - C) + D

A = Amplitude of the motion = Diameter/2 = 16/2 = 8

P = The period of the motion 2·π/B

B·x = The angle described by the motion = Half of one revolution = π = 180°

C = Phase shift = π

D = The midline = Diameter/2 = 8 inches

Therefore;

h = 8×cos(π - π) + 8 = 16 inches

After the bike moves forward another 56·π inches the height of the screw = 16 inches.

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2 years ago