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Ghella [55]
3 years ago
11

PLEASE HELP ME YOU GUYS ILL GIVE YOU A BRAINLIEST!!

Mathematics
2 answers:
Ulleksa [173]3 years ago
8 0

Answer:

(c) The graph starts flat but curves steeply upward.

Step-by-step explanation:

(a) The graph is a straight line that has a slope of 8

In the table, the value y increases. there is no constant increase in y

So The graph is not a straight line

(b) The graph is a horizontal line at y = 16

For y=16 graph, the value of y remains the same

The value of y changes in the table

So the graph is not a horizontal line

(c) The graph starts flat but curves steeply upward.

The value of y on the table increases rapidly. So we can say that graph starts flat and curves steeply upward.

(d) The graph is a parabola that opens upward.

There is no vertex mentioned in the table. the y values goes on increasing.

In parabola, the graph increases and then decreases or vice versa .

So the graph of given table is not a parabola


AVprozaik [17]3 years ago
7 0
The graph starts flat but then curves steeply upwards. You can tell this by the sudden jumps in y-coordinates, illustrating that it keeps going further up faster and faster as the x-coordinates progress steadily. 
You might be interested in
Show that if X is a geometric random variable with parameter p, then
Lubov Fominskaja [6]

Answer:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

P(X=x)=(1-p)^{x-1} p

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

X\sim Geo (1-p)

In order to find the expected value E(1/X) we need to find this sum:

E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}

Lets consider the following series:

\sum_{k=1}^{\infty} b^{k-1}

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}   (a)

On the last step we assume that 0\leq r\leq b and \sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}, then the integral on the left part of equation (a) would be 1. And we have:

\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)

And for the next step we have:

\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}

And with this we have the requiered proof.

And since b=1-p we have that:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

4 0
3 years ago
Help pls will mark branliest !
Mnenie [13.5K]

Answer:

10 Numbers

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
f(x) = 2<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class="latex
loris [4]

Answer:

No answer is possible

Step-by-step explanation:

First, we can identify what the parabola looks like.

A parabola of form ax²+bx+c opens upward if a > 0 and downward if a < 0. The a is what the x² is multiplied by, and in this case, it is positive 2. Therefore, this parabola opens upward.

Next, the vertex of a parabola is equal to -b/(2a). Here, b (what x is multiplied by) is 1 and a =2, so -b/(2a) = -1/4 = -0.25.

This means that the parabola opens upward, and is going down until it reaches the vertex of x=-0.25 and up after that point. Graphing the function confirms this.

Given these, we can then solve for when the endpoints of the interval are reached and go from there.

The first endpoint in -2 ≤ f(x) ≤ 16 is f(x) = 2. Therefore, we can solve for f(x)=-2 by saying

2x²+x-4 = -2

add 2 to both sides to put everything on one side into a quadratic formula

2x²+x-2 = 0

To factor this, we first can identify, in ax²+bx+c, that a=2, b=1, and c=-2. We must find two values that add up to b=1 and multiply to c*a = -2  * 2 = -4. As (2,-2), (4,-1), and (-1,4) are the only integer values that multiply to -4, this will not work. We must apply the quadratic formula, so

x= (-b ± √(b²-4ac))/(2a)

x = (-1 ± √(1-(-4*2*2)))/(2*2)

= (-1 ± √(1+16))/4

= (-1 ± √17) / 4

when f(x) = -2

Next, we can solve for when f(x) = 16

2x²+x-4 = 16

subtract 16 from both sides to make this a quadratic equation

2x²+x-20 = 0

To factor, we must find two values that multiply to -40 and add up to 1. Nothing seems to work here in terms of whole numbers, so we can apply the quadratic formula, so

x = (-1 ± √(1-(-20*2*4)))/(2*2)

= (-1 ± √(1+160))/4

= (-1 ± √161)/4

Our two values of f(x) = -2 are (-1 ± √17) / 4 and our two values of f(x) = 16 are (-1 ± √161)/4 . Our vertex is at x=-0.25, so all values less than that are going down and all values greater than that are going up. We can notice that

(-1 - √17)/4 ≈ -1.3 and (-1-√161)/4 ≈ -3.4 are less than that value, while (-1+√17)/4 ≈ 0.8 and (-1+√161)/4 ≈ 2.9 are greater than that value. This means that when −2 ≤ f(x) ≤ 16 , we have two ranges -- from -3.4 to -1.3 and from 0.8 to 2.9 . Between -1.3 and 0.8, the function goes down then up, with all values less than f(x)=-2. Below -3.4 and above 2.9, all values are greater than f(x) = 16. One thing we can notice is that both ranges have a difference of approximately 2.1 between its high and low x values. The question asks for a value of a where a ≤ x ≤ a+3. As the difference between the high and low values are only 2.1, it would be impossible to have a range of greater than that.

7 0
2 years ago
The proportion of students at a college who have GPA higher than 3.5 is 19%. a. You take repeated random samples of size 25 from
melomori [17]

Answer:

\mu_{\hat{p}}=0.19

\sigma_{\hat{p}}=0.0785

Step-by-step explanation:

We know that the mean and the standard error of the sampling distribution of the sample proportions will be :-

\mu_{\hat{p}}=p

\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}

, where p=population proportion and n= sample size.

Given : The proportion of students at a college who have GPA higher than 3.5 is 19%.

i.e. p= 19%=0.19

The for sample size n= 25

The mean and the standard error of the sampling distribution of the sample proportions will be :-

\mu_{\hat{p}}=0.19

\sigma_{\hat{p}}=\sqrt{\dfrac{0.19(1-0.19)}{25}}\\\\=\sqrt{0.006156}=0.0784601809837\approx0.0785

Hence , the mean and the standard error of the sampling distribution of the sample proportions :

\mu_{\hat{p}}=0.19

\sigma_{\hat{p}}=0.0785

8 0
3 years ago
What is the sum of series is numbers 64&amp;66 ?<br> Using the summation notation
iragen [17]

Answer:

64. 6138

66. 8.078125

Step-by-step explanation:

Let's take it one number at a time.

For number 64:

6(2)^{1-1}=6

6(2)^{2-1}=12

6(2)^{3-1}=24

6(2)^{4-1}=48

6(2)^{5-1}=96

6(2)^{6-1}=192

6(2)^{7-1}=384

6(2)^{8-1}=768

6(2)^{9-1}=1536

6(2)^{10-1}=3072

Sum = 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 + 1536 + 3072

Sum = 6138

For number 66:

12(-\frac{1}{2})^{0}=12

12(-\frac{1}{2})^{1}=-6

12(-\frac{1}{2})^{2}=3

12(-\frac{1}{2})^{3}=-1.5

12(-\frac{1}{2})^{4}=0.75

12(-\frac{1}{2})^{5}=-0.375

12(-\frac{1}{2})^{6}=0.1875

12(-\frac{1}{2})^{7}=-0.0078125

12(-\frac{1}{2})^{8}=0.046875

12(-\frac{1}{2})^{9}=-0.0234375

Sum = 12 + -6 + 3 + -1.5 + 0.75 + -0.375 + 0.1875 + -0.0078125 + 0.046875 + -0.0234375

or to make things simpler:

Sum = 12 - 6 + 3 - 1.5 + 0.75 - 0.375 + 0.1875 - 0.0078125 + 0.046875 - 0.0234375

Sum = 8.078125

4 0
3 years ago
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