PLEASE ANSWER THE QUESTION

Find the Domain and range of each relation. 1. Q= {(3,4), (-4,1), (2,5), (-4,1), (0,0)}

zimovet [89]

Answer:

Step-by-step explanation:

domain is all ur x values.....

domain = { -4,0,2,3}....u only have to write them once if they repeat

range is all ur y values....

range = { 0,1,4,5}

7 0

3 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co

Ira Lisetskai [31]

Answer:

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 103-1 = 102$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have $T = 2.63$ .

Now, we need to find the standard deviation of the sample. That is:

$s = \frac{0.56}{\sqrt{103}} = 0.055$

Now, we multiply T and s

$M = T*s = 2.63*0.055 = 0.145$

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

5 0

3 years ago

F(t) = t-1 g(t)= t^3 + 4t Find (f -g)(t)

Annette [7]

Answer:

Step-by-step explanation:

t - 1 - t^3 - 4t

-t^3 - 3t

5 0

3 years ago

Find an equation of the parabola y = ax2 + bx + c that passes through (0, 1) and is tangent to the line y = 4x − 4 at (1, 0).

larisa86 [58]

We are given the following:
- parabola passes to both (1,0) and (0,1)
 - slope at x = 1 is 4 from the equation of the tangent line 

First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. When x = 0, y = 1. So, c should be equal to 1. The parabola is y = ax^2 + bx + 1 

Now, we can substitute the point (1,0) into the equation,
0 = a(1)^2 + b(1) + 1
0 = a + b + 1
a + b = -1 

The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.
We take the derivative of the equation ,
y = ax^2 + bx + 1
y' = 2ax + b

x = 1, y' = 2
4 = 2a(1) + b
4 = 2a + b 

So, we have two equations and two unknowns, 
2a + b = 4 
a + b = -1

Solving simultaneously,
a = 5 
b = -6

Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .

6 0

3 years ago

Number puzzle:

ra1l [238]

6 times (7+5) plus 3 or 6x(7+5)+3

3 0

3 years ago

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