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marissa [1.9K]
2 years ago
6

A basketball player dropped a ball from a height of 5.5 feet. The first bounce reached a height of approximately 16 inches. The

second bounce reached 4 inches. What exponential function would be a good model to represent maximum ball height as a function of the number of bounces?
Mathematics
1 answer:
DerKrebs [107]2 years ago
7 0

Answer:

y=66(.25)x

(obviously the x is supposed to be an exponent but idk how to make it an exponent)

Step-by-step explanation:

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Elisa puts $1,000 into each of 2 bank accounts.
Dafna1 [17]

Answer:

How much money will be in Account A at the end of 3 years? $ ⇒ 1092.73

How much money will be in Account B at the end of 3 years? $112

Step-by-step explanation:

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2 years ago
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An ethanol railroad tariff is a fee charge for shipments of ethanol on public railroads. The Agriculture Marketing Service publi
Stels [109]

Answer:

The appropriate response is "0.9476".

Step-by-step explanation:

The given values are:

Standard deviation,

\sigma=1150

Randomly selected railroad-car shipments,

n = 500

As we know,

⇒ \sigma \bar{x}=\frac{\sigma}{\sqrt{n} }

On substituting the values, we get

⇒      =\frac{1150}{\sqrt{500} }

then,

⇒      =P(-1.94

⇒      =P(Z

On using the table of P, we get

⇒      =0.9738 - 0.0262

⇒      =0.9476

3 0
3 years ago
Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow
prisoha [69]

Answer:

The answers is " Option B".

Step-by-step explanation:

CI=\hat{Y}\pm t_{Critical}\times S_{e}

Where,

\hat{Y}= predicted value of lead content when traffic flow is 15.

\to df=n-1=8-1=7

 95\% \ CI\  is\  (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\

     =\frac{1059.8}{2}\\\\=529.9

Calculating thet-critical valuet_{ \{\frac{\alpha}{2},\ df \}}=-2.365

The lower predicted value =529.9-2.365(Se)

463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076

When 99\% of CI use as the expected lead content: \to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)

8 0
2 years ago
Did I do this right?
Rasek [7]

Answer:

Yea u did......

Step-by-step explanation:

you did right because all the steps are right

5 0
2 years ago
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The masses mi are located at the points Pi. Find the moments Mx and My and the center of mass of the system. m1 = 4, m2 = 3, m3
viva [34]

Answer:

M_x= 8

M_y = 6

Therefore the co-ordinate of the center of mass is = (\frac{4}{5},\frac{3}{5})

Step-by-step explanation:

Center of mass: Center of mass of an object is a point on the object. Center of mass is the average position of the system.

Center of mass of a triangle is the centriod of a triangle.

Given that m₁= 4, m₂=3, m₃=3 and the points are P₁(2,-3), P₂(-3,1) and P₃(3,5)

M_x = ∑(mass × x-co-ordinate)

M_y = ∑(mass × y-co-ordinate)

Therefore  

M_x = (4×2)+{3×(-3)}+(3×3)

     =8

M_y = {4×(-3)}+{3×1}+(3×5)

    =6

The x co-ordinate of the center of mass is the ratio of M_x to the total mass.

The y co-ordinate of the center of mass is the ratio of M_y to the total mass.

Total mass (m) = m₁+ m₂+ m₃

                        = 4+3+3

                        =10

The x co-ordinate of the center of mass is \frac {8}{10} = \frac {4}{5}

The y co-ordinate of the center of mass is \frac{6}{10}=\frac{3}{5}

Therefore the co-ordinate of the center of mass is = (\frac{4}{5},\frac{3}{5})

4 0
3 years ago
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