Looking for the Domain and range

What is the axis of a graph used for? Select all that apply. a starting point with equal intervals that follow a stopping point

Ronch [10]

Answer:

A scale to plot data

It is hard to tell the difference between the choices. If they are the following:

a starting point with equal intervals that follow
a stopping point for the data that can fit on the graph
a way to locate data
a scale to plot data

3 0

3 years ago

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Suppose the manager of a gas station monitors how many bags of ice he sells daily along with recording the highest temperature e

Maslowich

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Equation of regression line :

Yˆ = −114.05+2.17X

X = Temperature in degrees Fahrenheit (°F)

Y = Number of bags of ice sold

On one of the observed days, the temperature was 82 °F and 66 bags of ice were sold.

X = 82°F ; Y = 66 bags of ice sold

1. Determine the number of bags of ice predicted to be sold by the LSR line, Yˆ, when the temperature is 82 °F.

X = 82°F

Yˆ = −114.05+2.17(82)

Y = - 114.05 + 177.94

Y = 63.89

Y = 64 bags

2. Compute the residual at this temperature.

Residual = Actual value - predicted value

Residual = 66 - 64 = 2 bags of ice

4 0

3 years ago

On your next turn in a board game there is a 60% chance that you will move ahead 4 spaces and a 40% chance that you will move ba

Vika [28.1K]

The expectation value is simply obtained by multiplying the probability of each outcome by each possible value. These are then summed for each outcome to obtain the expectation value. In the board game, if there is a 60% chance that you will move 4 spaces forward (+4) and a 40% chance that you will move back 2 spaces then expected value for number of spaces is (0.6 x 4) + (0.4 x -2) = 2.4 - 0.8 = 1.6 spaces.

8 0

3 years ago

Can someone please help me out?

jek_recluse [69]

Answer:

Ligh blue

Step-by-step explanation:

6 0

3 years ago

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Sarah claims that the thickness of the spearmint gum she produces is 7.5 one-hundredths of an inch. A quality control specialist

satela [25.4K]

Answer:

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

$p_v =2*P(t_{(9)}>1.539)=0.158$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis.

Step-by-step explanation:

Data given and notation

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And the sample deviation given by:

$s=\sart{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And we got:

$\bar X=7.55$ represent the sample mean

$s=0.103$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =7.5$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 7.5 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 7.5$

Alternative hypothesis: $\mu \neq 7.5$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=10-1=9$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(9)}>1.539)=0.158$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis.

3 0

3 years ago