Can anyone help us on this question

An amusement park studied methods for decreasing the waiting time (minutes) for rides by loading and unloading riders more effic

Digiron [165]

Answer:

see explaination

Step-by-step explanation:

The p-value for Factor A is greater than 0.1

What is your conclusion with respect to Factor A?

Factor A is not significant.

The p-value for Factor B is greater than 0.1

What is your conclusion with respect to Factor B?

Factor B is not significant.

The p-value for the interaction of factors A and B is greater than 0.1

What is your conclusion with respect to the interaction of Factors A and B?

The interaction of Factors A and B is not significant.

What is your recommendation to the amusement park?

Since method is not a significant factor, use either loading and unloading method.

4 0

3 years ago

Vhich expression is equivalent to 16^3

LenKa [72]

Answer:16×16×16

Step-by-step explanaution:

16 ^3 so that means you need to but 16 3 times

6 0

2 years ago

Jeff has a density of 1000 g/mL and a volume of 25-mL. What is Jeff’s mass?

zlopas [31]

Answer:

25,000 grams

Step-by-step explanation:

Density can be found using the following formula:

$d=\frac{m}{v}$

If we rearrange the formula for $m$ , mass, we will multiply both sides of the equation by v.

$d*v=\frac{m}{v}*v$

$d*v=m$

The mass can be found by multiplying the density by the volume.

Jeff's density is 1000 grams per milliliter and his volume is 25 milliliters.

$d= 1000 g/mL\\v=25 mL$

Substitute the values into the formula.

$m=d*v$

$m= 1000 g/mL * 25 mL$

Multiply. When you multiply, the "mL" or milliliters will cancel out.

$m= 1000 g * 25$

$m= 25000 g$

Jeff's mass is 25,000 grams.

6 0

3 years ago

On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes

belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

$r = \frac{\Delta T_{X}}{\Delta T_{Y}}$

$r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}$

$r = 11\,\frac{^{\circ}X}{^{\circ}Y}$

The difference between current temperature in Y linear scale with respect to freezing point is:

$\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)$

$\Delta T_{Y} = 115\,^{\circ}Y$

The change in X linear scale is:

$\Delta T_{X} = r\cdot \Delta T_{Y}$

$\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)$

$\Delta T_{X} = 1265\,^{\circ}X$

Lastly, the current temperature on the X scale is:

$T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X$

$T_{X} = 1150\,^{\circ}X$

The current temperature on the X scale is 1150 °X.

5 0

2 years ago

Help please, I need it

Helen [10]

Partial Answer:

For #10 the solutions are 2 and 5

Step-by-step explanation:

Solutions for an equation can be x-intercepts, or where it touches the x or horizontal line. The equation in #10 touches the x line at 2 and 5.

5 0

2 years ago