A runner increases his velocity from 0 m/s to 20 m/s in 2.0 s. What was his average acceleration?

What are the domain and range of the graph? Is the graph a function? Explain.

kumpel [21]

Answer:

Domain: -3 < x <2

Range: -4 < y < 5

Is the graph a function? No the graph isn't not, since the graph fails the vertical line test

8 0

3 years ago

Samantha reports that 60% of the first year students at her university think they should be able to bring a car to campus. She a

Nostrana [21]

Answer:

D) the standard error for her sample is 15

Step-by-step explanation:

Samantha is 95% certain that between 50% and 70% of the first year students agree that they should be able to bring a car to campus.

Here

95% is the confidence level.
The range between 50% and 70% indicates 95% confidence interval.
60% is the average proportion of the first year students at her university think they should be able to bring a car to campus.
10% (0.1) is the margin of error from the mean

Since standard error can be obtained by dividing margin of error by the z-statistic of the confidence level which is 1.96. The standard error for her sample $\frac{0.1}{1.96}$ does not equal to 15.

4 0

3 years ago

Ray wants to buy the larger of two aquariums. one aquarium has a base that is 20 inches by 20 inches and height that is 18 inche

Svetradugi [14.3K]

Ray should buy the first one because it has the greater volume by 1,440 square inches.
Tank 1: 20x20x18=7,200
Tank 2: 40x12x12=5,760
7,200-5,760=1,440

4 0

3 years ago

Read 2 more answers

Pls help me solve this please I will give 70 points please help

Mashutka [201]

Answer:

Fraction Form: $\frac{2187}{4000000}$

Decimal Form: $0.000547$

Step-by-step explanation:

$=\frac{5}{256} * \frac{3}{5}^1^2 / \frac{3}{5}^5$

$\frac{3}{5}^1^2/\frac{3}{5}^5=\frac{2187}{78125}$

$=\frac{5}{256}*\frac{2187}{78125}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}$

$\frac{5*2187}{256*78125}$

$\mathrm{Factor\:the\:number:\:}\:78125=5\cdot \:15625$

$=\frac{5*2187}{256*5*15625}$

$\mathrm{Cancel\:the\:common\:factor:}\:5$

$=\frac{2187}{256*15625}$

$\mathrm{Multiply\:the\:numbers:}\:256\cdot \:15625=4000000$

$\frac{2187}{4000000}$

[RevyBreeze]

5 0

2 years ago

People tend to evaluate the quality of their lives relative to others around them (Frieswijk et al., 2004). In one study, resear

Alborosie

Answer:

Step-by-step explanation:

Hello!

Research.

n=9 frail elderly were interview and compared to a fictitious person who was worse off then the interviewee, a life-satisfaction score was determined for each person.

18, 23, 24, 22, 19, 27, 23, 26, 25

Assuming that the population average score is μ= 20, the researchers think that the elderly in the sample are more or less satisfied than others in the general population.

a. You have the information of one sample, assuming this sample has a normal distribution and each elderly interviewed is independent, then the t-test of choice is a one-sample t-test.

b. and c. If you say that the elderly are "more or less" satisfied than the others, this means that they are either as satisfied as to the general population or not satisfied as to the general population. Symbolically:

H₀: μ = 20

H₁: μ ≠ 20

This is a two-tailed test, meaning, you will have two critical regions.

d.

α: 0.05

Left critical value: $t_{n-1;/\alpha 2} = t_{8; 0.025}= -2.306$

Right critical value: $t_{n-1;1-\alpha /2} = t_{8;0.975} = 2.306$

e.

$t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~t_{n-1}$

X[bar]= 23

S= 3

$t_{H_0}= \frac{23-20}{\frac{3}{\sqrt{9} } }= 3$

f.

Considering that the calculated t-value is greater than the right critical value, the decision is to reject the null hypothesis, so using a significance level of 5% you can conclude that the average life-satisfaction score of the elderly is different than 20.

I hope it helps!

7 0

3 years ago