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zvonat [6]
3 years ago
11

Math help, thanks for the help. i'm very great full

Mathematics
1 answer:
rodikova [14]3 years ago
4 0

4x +28 and 68 are vertical angles and are equal.

Set them to equal each other to solve for x:

4x +28 = 68

Subtract 28 from each side:

4x = 40

Divide both sides by 4:

x = 40/4

x = 10

You might be interested in
The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir
Paha777 [63]

Answer:

95​% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% 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 time = 5.15 years

            \sigma = sample standard deviation = 1.68 years

            n = sample of college graduates = 4400

            \mu = population mean time

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                               level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                              = [ 5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } } , 5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } } ]

                                             = [5.10 , 5.20]

Therefore, 95​% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

8 0
3 years ago
A buoy moves up and down with the waves in simple harmonic motion. This buoy moves a total of 12 feet from its minimum position
Lana71 [14]
That was rude of him
3 0
3 years ago
Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'
vodka [1.7K]

Answer:

a) v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

b)  0

c) a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

5 0
3 years ago
Find the perimeter of the triangle using the dimensions shown given below. (a = 57.36 yards; c = 83 yards; h = 54 yards) *
marishachu [46]

Answer:

I'm glad you asked!

Step-by-step explanation:

54 +83+57.36+83 = 211.42

7 0
3 years ago
For a history fair, a school is building a circular wooden stage that will stand that will stand 2 feet off the ground . Find th
yawa3891 [41]
To find the answer you do 19×19= 361 and do 361×3.14= 1,133.54 The area of the stage is 1,133.54
8 0
3 years ago
Read 2 more answers
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