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Miya went to the bowling alley. There was a $5 shoe rental fee plus $15.0 per game. Which equation models the situation where g

Aloiza [94]

Answer:

$15g + $5 = c

7 0

2 years ago

Javier made a scatterplot to show the data he collected on the growth of a plant.

Leya [2.2K]

Answer:

The answer is C i.e y = 3.25 x + 4.60

Step-by-step explanation:

Given the graph in which the Javier made a scatter plot to show the data he collected on the growth of a plant.

Now, we have to choose the equation which best represents Javier's data.

The graph shown does not pass through origin therefore intercept can not be equal to 0. hence solution A discarded.

The graph of rest of three solutions attached and the points which shown in the graph match to the points on the graph of solution third as shown. Hence, The answer is C i.e y = 3.25 x + 4.60

4 0

3 years ago

A fabric store has 80 3/8 yards of a particular fabric. About how many pairs of curtains could be made from this fabric if each

Keith_Richards [23]

20 because 19 16/33 is the exact answer. Since they ask about how much that can be rounded up to 20

4 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago

Turn 18 2/5 into a decimal

Studentka2010 [4]

18 2/5 = 92/5 = 18.4 as a decimal;

6 0

3 years ago