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True [87]
3 years ago
6

Is 8.25 lbs more than 8.225 lbs

Mathematics
1 answer:
den301095 [7]3 years ago
6 0

Answer:

yes

Step-by-step explanation:

8.25 would be since in the 0.05 would be larger than 0.02

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Jason tossed a fair coin 3 times. What is the probability of getting a head and two tails in any order 1.3/8 2.4/8 3.5/8 4.6/8 p
ladessa [460]

Answer:

1. 3/8

Step-by-step explanation:

Total tossed times = 3

Total Possibilities = 8

Possibilities = (Head, Head, Head), (Tails,Tail,Tails), (Head,Head,Tails), (Head,Tail,Head), (Tails,Head,Head), (Head,Tail,Tail), (Tail,Tail,Head), (Tail,Head,Tail)

Number of options with a head and two tails = 3

So, Probability = Number of correct options ÷ Number of total possibilities

= 3 ÷ 8 or 3/8

4 0
3 years ago
I need help due tmrw
White raven [17]

Answer:

yo ill help you but do you happen to have a argumentative essay written ill help u if you help me

3 0
2 years ago
Find the inverse of f(x)= 5x-2
scZoUnD [109]

Find the function by taking the integral of the derivative.

F(x)=52x2−2x+C

5 0
3 years ago
The CEO of a large corporation asks his Human Resource (HR) director to study absenteeism among its executive-level managers at
Fynjy0 [20]

Answer:

Following are the answer to this question:

Step-by-step explanation:

Given:

n = 30 is the sample size.  

The mean  \bar X = 7.3 days.  

The standard deviation = 6.2 days.  

df = n-1  

     = 30-1 \\      =29

The importance level is \alpha = 0.10  

The table value is calculated with a function excel 2010:

= tinv (\ probility, \ freedom \ level) \\= tinv (0.10,29) \\ =1.699127\\ =  t_{al(2x-1)}= 1.699127

The method for calculating the trust interval of 90 percent for the true population means is:

Formula:

\bar X - t_{al 2,x-1} \frac{S}{\sqrt{n}} \leq \mu \leq \bar X+ t_{al 2,x-1}   \frac{S}{\sqrt{n}}

=\bar X - t_{0.5, 29} \frac{6.2}{\sqrt{30}} \leq \mu \leq \bar X+ t_{0.5, 29}   \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 \frac{6.2}{\sqrt{30}}\leq \mu \leq7.3 +1.699127 \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 (1.13196)\leq \mu \leq7.3 +1.699127  (1.13196) \\\\=5.37 \leq \mu  \leq 9.22 \\

It can rest assured that the true people needs that middle managers are unavailable from 5,37 to 9,23 during the years.

7 0
2 years ago
Which statements are true about the graph of the function f(x) = x2 – 8x + 5? Check all that apply.
statuscvo [17]

Answer:

A, D, E are true

Step-by-step explanation:

You have to complete the square to prove A.  Do this by first setting the function equal to 0, then moving the 5 to the other side.

x^2-8x=-5

Now we can complete the square.  Take half the linear term, square it, and add it to both sides.  Our linear term is 8 (from the -8x).  Half of 8 is 4, and 4 squared is 16.  So we add 16 to both sides.

(x^2-8x+16)=-5+16

We will do the addition on the right, no big deal.  On the left, however, what we have done in the process of completing the square is to create a perfect square binomial, which gives us the h coordinate of the vertex.  We will rewrite with that perfect square on the left and the addition done on the right,

(x-4)^2=11

Now we will move the 11 back over, which gives us the k coordinate of the vertex.

(x-4)^2-11=y

From this you can see that A is correct.

Also we can see that the vertex of this parabola is (4, -11), which is why B is NOT correct.

The axis of symmetry is also found in the h value.  This is, by definition, a positive x-squared parabola (opens upwards), so its axis of symmetry will be an "x = " equation.  In the case of this type of parabola, that "x = " will always be equal to the h value.  So the axis of symmetry is

x = 4, which is why C is NOT correct, either.

We can find the y-intercept of the function by going back to the standard form of the parabola (NOT the vertex form we found by completing the square) and sub in a 0 for x.  When we do that, and then solve for y, we find that when x = 0, y = 5.  So the y-intercept is (0, 5).

From this you can see that D is also correct.

To determine if the parabola has real solutions (meaning it will go through the x-axis twice), you can plug it into the quadratic formula to find these values of x.  I just plugged the formula into my graphing calculator and graphed it to see that it did, indeed, go through the x-axis twice.  Just so you know, the values of x where the function go through are (.6833752, 0) and (7.3166248, 0).  That's why you need the quadratic formula to find these values.

7 0
3 years ago
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