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Nadusha1986 [10]
2 years ago
9

Two boats depart from a port located at (–10, 0) in a coordinate system measured in kilometers, and they travel in a positive x-

direction. the first boat follows a path that can be modeled by a quadratic function with a vertex at (0, 5), and the second boat follows a path that can be modeled by a linear function and passes through the point (10, 4). at what point, besides the common starting location of the port, do the paths of the two boats cross? (–6, 0.8) (–6, 3.2) (6, 3.2) (6, 0.8)
Mathematics
2 answers:
stiks02 [169]2 years ago
5 0
To get the points at which the two boats meet we need to find the equations that model their movement:
Boat A:
vertex form of the equation is given by:
f(x)=a(x-h)^2+k
where:
(h,k) is the vertex, thus plugging our values we shall have:
f(x)=a(x-0)^2+5
f(x)=ax^2+5
when x=-10, y=0 thus
0=100a+5
a=-1/20
thus the equation is:
f(x)=-1/20x^2+5

Boat B
slope=(4-0)/(10+10)=4/20=1/5

thus the equation is:
1/5(x-10)=y-4
y=1/5x+2

thus the points where they met will be at:
1/5x+2=-1/20x^2+5
solving for x we get:
x=-10 or x=6
when x=-10, y=0
when x=6, y=3.2
Answer is (6,3.2)
nexus9112 [7]2 years ago
4 0

The answer is C. (6,3.2) on Edg ;))

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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
jok3333 [9.3K]

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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2 years ago
Christopher wrote the number pattern below.The first term is 8. 8,6,8,7,10
Aneli [31]
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Osvoldo has a goal of getting at least 30\%30%30, percent of his grams of carbohydrates each day from whole grains. Today, he at
Alchen [17]
Hi there! Osvoldo did not meet his goal.

(In my answering I suppose you mean 30%, 220 grams of carbohydrates and 55 grams of whole grains, if this is incorrect, please let me know)

Today Osvoldo ate 220 grams of carbohydrates.
10 % of 220 is 22 (divide by 10), and therefore 30 % of 220 is 22 * 3 = 66. 

If at least 66 grams of Osvoldo's total consumption consisted of whole grains, he would have met his goal. However, he only ate 55 grams of whole grains (which is less than 66), and therefore he did not meet his goal.
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2 years ago
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1 year ago
g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the
blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean amount of money spent on textbooks = $121.60

            \sigma = population standard deviation = $6.36

            n = sample of students = 36

            \mu = population mean

<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<em />

So, 95% confidence interval for the population mean, \mu is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                      of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

                                      = [ 121.60-1.645 \times {\frac{6.36}{\sqrt{36} } } , 121.60+1.645 \times {\frac{6.36}{\sqrt{36} } } ]

                                      = [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0
2 years ago
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