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IRINA_888 [86]
3 years ago
15

(A) find F(x) for the indicated values of x, if possible

Mathematics
1 answer:
dem82 [27]3 years ago
8 0

Answer:

A) The indicated values of F(x) for x=-3 and x=5 are F(-3)=-27  and F(5)=125

B) The domain of f is the set of all real numbers

Step-by-step explanation:

Given that the function F is defined by

F(x)=x^3 for x=-3,5

A) To find F(x) for the indicated values of x :

Given F(x)=x^3 for x=-3,5

  • Put x=-3 in the given function

F(-3)=(-3)^3

=(-3)\times (-3)\times (-3)

=-27

Therefore F(-3)=-27

  • Put x=5 in the given function

F(5)=(5)^3

=(5)\times (5)\times (5)

=125

Therefore F(5)=125

The indicated values of F(x) for x=-3 and x=5 are F(-3)=-27  and F(5)=125

B) To find the domain of f :

The domain of the f in the given expression is the set of all real numbers except where the expression x^3 is undefined. In this case, there is no real number that makes the expression undefined.

The domain of f is the set of all real numbers

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Answer:

The data is skewed to the bottom and contains an outlier.  

Step-by-step explanation:

1. Test for outlier

An outlier is a point that is more than 1.5IQR below Q1 or above Q3.  

IQR = Q3 - Q1 = 74 - 51 = 23

1.5 IQR  = 1.5 × 23 = 34.5

51 - 15 = 36 > 1.5IQR

The point at 15 is an outlier.

2. Test for normal distribution

The median is not in the middle of the box.  

Rather, it cuts the box into two unequal parts, so the data does not have a normal distribution.

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Answer for branliest
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4 0
3 years ago
If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,
julsineya [31]
<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)

(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

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