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tiny-mole [99]
1 year ago
7

PLEASE HELP!!! ASAP The function graphed approximates the height of a rock, in meters, x seconds after it falls from a cliff

Mathematics
2 answers:
nasty-shy [4]1 year ago
8 0

The vertical numbers are the distance and the horizontal numbers are the time in seconds.

Look at where the line is located at the number 20.

It si between the 2 and the 3 on the horizontal line, but you can see it is closer to the 2 than it is the 3, so the time would be 2.2s

True [87]1 year ago
6 0

Answer:

2.2 seconds

Step-by-step explanation:

The attached graph shows the height of a rock after it falls from a cliff. Let after t seconds the rock is 20 meters from the ground.

If the graph is not given we can calculate the time using the second equation of motion i.e.

d=ut+\dfrac{1}{2}at^2

Where

u is the initial speed of the rock

a is the acceleration of the rock

t is the time taken

Here, distance covered is in y axis and the time taken is in x axis. Approximately, at 2.2 seconds the rock is 20 meters above the ground. Hence, this is the required solution.

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OLga [1]
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3 0
1 year ago
A web-based company has a goal of processing 90 percent of its orders on the same day they are received. If 434 out of the next
Kamila [148]

Answer:

We conclude that a web-based company are not exceeding their goal of 90%.

Step-by-step explanation:

We are given that a web-based company has a goal of processing 90 percent of its orders on the same day they are received.

434 out of the next 471 orders are processed on the same day.

Let p = <u><em>proportion of orders processing on the same day they are received.</em></u>

SO, Null Hypothesis, H_0 : p \leq 0.90     {means that they are not exceeding their goal of 90%}

Alternate Hypothesis, H_0 : p > 0.90      {means that they are exceeding their goal of 90%}

The test statistics that would be used here <u>One-sample z test for</u> <u>proportions</u>;

                            T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of orders that are processed on the same day = \frac{434}{471} = 0.92

           n = sample of orders = 471

So, <u><em>the test statistics</em></u>  =  \frac{0.92-0.90}{\sqrt{\frac{0.92(1-0.92)}{471} } }

                                     =  1.599

The value of z test statistics is 1.599.

<u>Now, at 0.025 significance level the z table gives critical value of 1.96 for right-tailed test.</u>

Since our test statistic is less than the critical value of z as 1.599 < 1.96, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that a web-based company are not exceeding their goal of 90%.

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1 year ago
Brian irons \dfrac29 9 2 ​ start fraction, 2, divided by, 9, end fraction of his shirt in 3\dfrac353 5 3 ​ 3, start fraction, 3,
Artyom0805 [142]
For this case we can make the following rule of three:
 2/9 ------> 3/5
 x ---------> 1
 Clearing x we have:
 x = (1 / (3/5)) * (2/9)
 Rewriting we have:
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he irons 10/27 of his shirt every minute
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1 year ago
Read 2 more answers
In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.
WARRIOR [948]

Answer:

For First Solution: y_1(t)=e^t

y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:y_2(t)=cosht

y_2(t)=cosht  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: y_1(t)=e^t

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_1(t)=e^t

First order derivative:

y'_1(t)=e^t

2nd order Derivative:

y''_1(t)=e^t

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:

y_2(t)=cosht

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_2(t)=cosht

First order derivative:

y'_2(t)=sinht

2nd order Derivative:

y''_2(t)=cosht

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence y_2(t)=cosht  is the solution of equation y''-y=0.

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Answer: 3.00

Step-by-step explanation:

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