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Darya [45]
3 years ago
12

An engineer is building a bridge that should be able to hold a maximum weight of 1 ton. He builds a model of the bridge that is

exactly 4 times smaller than the actual bridge.
16 ounces = 1 pound. 2,000 pounds = 1 ton.

If a test of the model shows that it holds 8,000 ounces, will the bridge hold 1 ton?

8,000 ounces on the model is equal to
ounces on the actual bridge.
Convert ounces to pounds. The actual bridge can hold
pounds.
Therefore, the bridge
hold 1 ton.
Mathematics
2 answers:
valentinak56 [21]3 years ago
7 0

Answer:

owa owa

Step-by-step explanation:

Sedaia [141]3 years ago
4 0

Answer:

the bridge holds one ton. on Edgunuity

Step-by-step explanation:     <3

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The height h(n) of a bouncing ball is an exponential function of the number n of bounces.
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Answer:

The height of a bouncing ball is defined by h(n) = 6\cdot \left(\frac{4}{6} \right)^{n-1}.

Step-by-step explanation:

According to this statement, we need to derive the expression of the height of a bouncing ball, that is, a function of the number of bounces. The exponential expression of the bouncing ball is of the form:

h = h_{o}\cdot r^{n-1}, n \in \mathbb{N}, 0 < r < 1 (1)

Where:

h_{o} - Height reached by the ball on the first bounce, measured in feet.

r - Decrease rate, no unit.

n - Number of bounces, no unit.

h - Height reached by the ball on the n-th bounce, measured in feet.

The decrease rate is the ratio between heights of two consecutive bounces, that is:

r = \frac{h_{1}}{h_{o}} (2)

Where h_{1} is the height reached by the ball on the second bounce, measured in feet.

If we know that h_{o} = 6\,ft and h_{1} = 4\,ft, then the expression for the height of the bouncing ball is:

h(n) = 6\cdot \left(\frac{4}{6} \right)^{n-1}

The height of a bouncing ball is defined by h(n) = 6\cdot \left(\frac{4}{6} \right)^{n-1}.

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There are 4 green marbles and 2 red marbles in the jar. You just randomly draw one by one without replacement and stop when you
zalisa [80]

Answer:

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3)  = 2/15

P(4) = 1/15

The expected value for X is:

EV = 1.33...

Step-by-step explanation:

We have a total of 6 marbles in the jar.

The probability of getting a red marble in the first try  (X = 0) is equal to the quotient between the number of red marbles and the total number of marbles, this is:

P(0) = 2/6 = 1/3

The probability of drawing one green marble (X = 1)

is:

First, you draw a green marble with a probability of 4/6

Then you draw the red one, but now there are 5 marbles in the jar (2 red ones and 3 green ones), then the probability is 2/5

The joint probability is:

P(1) = (4/6)*(2/5) = (2/3)*(2/5) = 4/15

The probability of drawing two green marbles (X  = 2)

Again, first we draw a green marble with a probability of 4/6

Now we draw again a green marble, now there are 3 green marbles and 5 total marbles in the jar, so this time the probability is 3/5

Now we draw the red marble (there are 2 red marbles and 4 total marbles in the jar), with a probability of 2/4

The joint probability is:

P(2) = (4/6)*(3/5)*(2/4) = (2/6)*(3/5) = 1/5

The probability of drawing 3 green marbles (X = 3)

At this point you may already understand the pattern:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

finally, we draw a red marble with a probability 2/3

The joint probability is:

P(3) = (4/6)*(3/5)*(2/4)*(2/3) = (2/6)*(3/5)*(2/3) = (1/5)*(2/3) = (2/15)

Finally, the probability of drawing four green marbles (X = 4) is given by:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

fourth, we draw a green marble with a probability 1/3

Finally, we draw a red marble with a probability 2/2 = 1

The joint probability is:

P(4) = (4/6)*(3/5)*(2/4)*(1/3)*1 = (1/5)*(1/3) = 1/15

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3)  = 2/15

P(4) = 1/15

The expected value will be:

EV = 0*P(0) + 1*P(1) + 2*P(2) + 3*P(3) + 4*P(4)

EV = 1*(4/15) + 2*( 1/5) + 3*( 2/15) + 4*(1/15 ) = 1.33

So we can expect to draw 1.33 green marbles in this experiment.

5 0
3 years ago
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