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Nesterboy [21]
1 year ago
10

a golfer needs to hit a ball a distance of 500 feet, but there is a 60-foot tall tree that is 100 feet in front of the point whe

re the shot needs to land. given that the maximum height of the shot is 120 feet, and that the intended distance of 500 feet is reached, by how much did the ball clear the tree?
Mathematics
1 answer:
jok3333 [9.3K]1 year ago
8 0

We can create a parabola equation of the trajectory using the vertex form:

y = a (x – h)^2 + k

 

The center is at h and k, where h and k are the points at the maximum height so:

h = 250

k = 120

 

Therefore:

 y = a (x – 250)^2 + 120

 

At the initial point, x = 0, y = 0, so we can solve for a:

0 = a (0 – 250)^2 + 120

0 = a (62,500) + 120

a = -0.00192

 

So the whole equation is:

y = -0.00192 (x – 250)^2 + 120

 

So find for y when the golf ball is above the tree, x = 400:

y = -0.00192 (400 - 250)^2 + 120

y = 76.8 ft

 

So the ball cleared the tree by:

76.8 ft – 60 ft = 16.8 ft

 

 

Answer:

16.8 ft

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Brilliant_brown [7]
Bala had 9 stickers

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3 0
2 years ago
A new curing process developed for a certain type of cement results in a mean compressive strength of 5000 kilograms per square
Sedbober [7]

Answer:

\alpha =0.0668

Step-by-step explanation:

Data given and notation  

The info given by the problem is:

n=25 the random sample taken

\mu =5000 represent the population mean

\sigma =100 represent the population standard deviation

The critical region on this case is \bar X so then if the value of \bar X \geq 4970 we fail to reject the null hypothesis. In other case we reject the null hypothesis

Null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the true mean is 5000, the system of hypothesis would be:  

Null hypothesis:\mu = 5000  

Alternative hypothesis:\mu \neq 5000  

Let's define the random variable X ="The compressive strength".

We know from the Central Limit Theorem that the distribution for the sample mean is given by:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

Find the probability of committing a type I error when H0 is true.

The definition for type of error I is reject the null hypothesis when actually is true, and is defined as \alpha the significance level.

So we can define \alpha like this:

\alpha= P(Error I)= P(\bar X

And in order to find this probability we can use the Z score given by this formula:

Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And the value for the probability of error I is givn by:

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1 year ago
If each lap in a pool is 100 meters long, how many laps equal one mile? Round to the nearest tenth. (Hint: 1 foot ≈0.3048 meter)
aalyn [17]
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bulgar [2K]

Answer:

a. Mean doesn't change.

b. Median doesn't change.

c. Mode can change.

Step-by-step explanation:

Let us assume the data set with 10 observations

{2,6,4,3,2,6,4,9,4,7}.

Arranging data set in ascending order

{2,2,3,4,4,4,6,6,7,9}

mean=2+2+3+4+4+4+6+6+7+9/10=4.7

median

n/2=10/2=5 is an integer so,

median= average of n/2 and n/2 +1

median= (5th value+6th value)/2

median=(4+4)/2=8/2=4

Mode

The most repeated value is 4. So, mode is 4 for assumed data.

Increasing highest value by 10 and decreasing lowest value by 10

{-8,2,3,4,4,4,6,6,7,19}

a.

mean=-8+2+3+4+4+4+6+6+7+19/10=4.7

Mean doesn't change

b.

median

n/2=10/2=5 is an integer so,

median= average of n/2 and n/2 +1

median= (5th value+6th value)/2

median=(4+4)/2=8/2=4

Median doesn't change

c.

Most occurring value is still 4. But mode can change if the value the highest value becomes most concurring value.

5 0
2 years ago
Read 2 more answers
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