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olchik [2.2K]
2 years ago
5

The base of the isosceles triangle is parallel to x-axis and has both end-points on the parabola y=x(10−x) and its vertex belong

s to the x-axis. Find area of the triangle if the length of its base is 8.
Mathematics
1 answer:
baherus [9]2 years ago
6 0

Answer:

36

Step-by-step explanation:

We must determine the x and y intercepts of the parabola:

When y=0, x=0 or x=10

WE know that the point of the triangle base is x and x+8. We can substitute this into the parabola equation because the endpoints are on the parabola.

f(x+8)=-(x^2+16x+64)+10x+80

f(x+8)=-x^2-6x+16

f(x)=f(x+8)

-x^2+10x=-x^2-6x+16

solve for x

16x=16

x=1

Therefore the heigh is f(1):

=-1^2+10=9

The area of the triangle is 1/2 base x height:

=(1/2)\cdot{8}\cdot{9}=36

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i = p \times r \times t
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Micah rows his boat on a river 4.48 miles downstream, with the current, in 0.32 hours. He rows back upstream the same distance,
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Let the speed of the current be y and the speed of Micah's sailing speed be x. Then 4.48/(x + y) = 0.32
4.48/(x - y) = 0.56

0.32x + 0.32y = 4.48 . . . (1)
0.56x - 0.56y = 4.48 . . . (2)
(1) x 7 => 2.24x + 2.24y = 31.36 . . . (3)
(2) x 4 => 2.24x - 2.24y = 17.92 . . . (4)
(3) - (4) => 4.48y = 13.44
y = 3
From (1), 0.32x + 0.32(3) = 4.48
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x = 3.52/0.32 = 11

Therefore, the speed of the current is 3 miles per hour.

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The average life of a bread-making machine is 7 years, with a standard deviation of 1 year. Assuming that the lives of these mac
Alina [70]

Answer:

a) P(6.4

b) a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable life of a bread making machine. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =7,\sigma =1)

We take a sample of n=9 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

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

\bar X \sim N(\mu=7, \frac{1}{\sqrt{9}})

Solution to the problem

Part a

(a) the probability that the mean life of a random sample  of 9 such machines falls between 6.4 and 7.2

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

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

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{1}{\sqrt{9}}=0.333

We want this probability:

P(6.4

Part b

b) The value of x to the right of which 15% of the  means computed from random samples of size 9 would fall.

For this part we want to find a value a, such that we satisfy this condition:

P(\bar X>a)=0.15   (a)

P(\bar X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.036. On this case P(Z<1.036)=0.85 and P(Z>1.036)=0.15

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.036

And if we solve for a we got

a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

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Answer:

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Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = T_1 = 15°c

The final temperature = T_2  = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q =  m × s × ( T_2 - T_1 )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q =  50 g × 2260.0 J/g × 85°c

∴   Q = 9,605,000  joule

Or, Q =  9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

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