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Kipish [7]
3 years ago
10

The table represents the height of a rock that is dropped, h(t), in meters, after t seconds.

Mathematics
2 answers:
laiz [17]3 years ago
5 0

Answer:

<h2>The rock hits the ground between 2 and 2.5 seconds.</h2>

Step-by-step explanation:

The given table is:

t            h(t)

0           20

0.5        18.8

1             15.1

1.5          9

2            0.4

2.5        -10.6

3           -24.1

The ground level is when h(t) = 0, which is between 0.4 and -10.6, because in that interval is included the zero. So, the problem is asking when does the rock hit the ground, referring to time. We just have to look for t-values for 0.4 and -10.6.

<em>Therefore, the rock hits the ground between 2 and 2.5 seconds.</em>

harina [27]3 years ago
4 0

Answer:

The rock hits the ground between 2

seconds and 2.5

seconds after it is dropped.

Step-by-step explanation:

At 2 seconds, the height of the rock is 0.4 meters, and at 2.5 seconds the height of the rock is -10.6 meters. Therefore, the rock hit the ground, or 0 meters, between those times.

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Answer:

b

Step-by-step explanation:

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3 years ago
A person operating a machine can mow 0.75 acres in ½ hour. What is the rate, in acres per hour, that the person can mow? Write y
Gekata [30.6K]

1.50 acres is the answer.

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3 years ago
A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted is incorrectly received wit
m_a_m_a [10]

Answer:

The probability that the message will be wrong when decoded is 0.05792

Step-by-step explanation:

Consider the provided information.

To reduce the chance or error, we transmit 00000 instead of 0 and 11111 instead of 1.

We have 5 bits, message will be corrupt if at least 3 bits are incorrect for the same block.

The digit transmitted is incorrectly received with probability p = 0.2

The probability of receiving a digit correctly is q = 1 - 0.2 = 0.8

We want the probability that the message will be wrong when decoded.

This can be written as:

P(X\geq3) =P(X=3)+P(X=4)+P(X=5)\\P(X\geq3) =\frac{5!}{3!2!}(0.2)^3(0.8)^{2}+\frac{5!}{4!1!}(0.2)^4(0.8)^{1}+\frac{5!}{5!}(0.2)^5(0.8)^0\\P(X\geq3) =0.05792

Hence, the probability that the message will be wrong when decoded is 0.05792

4 0
2 years ago
When n = 3, which of the expressions below will be greater than 50? Select all that apply. A) (8n2÷9)⋅7 B) 2n2⋅3 –4n C) 4n2–6n+1
True [87]

Answer:

<u><em>A</em></u>

Step-by-step explanation:

<u>a)</u>

<u></u>(8(3)^{2} / 9) * 7<u></u>

<u>8(9) / 9 = 8 x 7 = 56</u>

b)

2(3)^{2} * 3 - 4(3)

2(9) * 3 - 12

18 * 3 = 54 - 12 = 42

c)

4(3)^{2} - 6(3) + 12

4(9) - 18 + 12

36 - 18 = 18 + 12 = 30

d)

7(3) + (3 * 8)

21 + 24 = 45

5 0
3 years ago
PLEASE HELP!!! Given the functions, f(x) = 6x + 2 and g(x) = x - 7, perform the indicated operation. When applicable, state the
Brums [2.3K]

Answer:

The domain restriction for (f/g)(x) is x=7

Step-by-step explanation:

we have

f(x)=6x+2

g(x)=x-7

so

(f/g)(x)=\frac{6x+2}{x-7}

Remember that

the denominator can not be equal to zero

so

Find the domain restriction

x-7=0

x=7

therefore

The domain is all real numbers except the number 7

(-∞,7)∪(7,∞)

4 0
3 years ago
Read 2 more answers
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