What is the name of the polygon? please help

g100num [7]

The answer would be:
-2(x+1)

3 0

3 years ago

Directions for questions 4 & 5: We selected a random sample of 100 StatCrunchU students, 67 females and 33 males, and analyz

GaryK [48]

Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender Mean Std. dev. n

Female 2577.75 1916.29 67

Male 3809.42 2379.47 33

We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

$M_d=M_1-M_2=2577.75-3809.42=-1231.67$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795$

The t-value for a 95% confidence interval is t=1.96.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.96 \cdot 475.795=932.54$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = -1231.67-932.54=-2164.21\\\\UL=M_d+t \cdot s_{M_d} = -1231.67+932.54=-299.13$

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

6 0

3 years ago

A right triangle has side lengths a,b, and c as shown below. Use these lengths to find sin x, tan x, and cos x.

My name is Ann [436]

Step-by-step explanation:

sinx = a/b

tanx= a/c

cosx= c/b

6 0

3 years ago

Can someone please help? I will give thanks and 5 stars, and if possible, brainlist, if you explain how to solve the question.

Sav [38]

Answer:

1/14

Step-by-step explanation:

Odd numbers: 5, 7, 9

3/7 × 1/6

1/14

4 0

3 years ago

How do you solve for the slope of a line with the equation ax^2 + bx + c?

ololo11 [35]

<h2>The graph of y = ax^2 + bx + c </h2><h2>A nonlinear function that can be written on the standard form </h2><h2>ax2+bx+c,where a≠0 </h2><h2>All quadratic functions has a U-shaped graph called a parabola. The parent quadratic function is </h2><h2> y=x2 </h2><h2> The lowest or the highest point on a parabola is called the vertex. The vertex has the x-coordinate </h2><h2>x=−b2a </h2><h2>The y-coordinate of the vertex is the maximum or minimum value of the function. </h2><h2>a > 0 parabola opens up minimum value </h2><h2>a < 0 parabola opens down maximum value </h2><h2> A rule of thumb reminds us that when we have a positive symbol before x2 we get a happy expression on the graph and a negative symbol renders a sad expression. </h2><h2>The vertical line that passes through the vertex and divides the parabola in two is called the axis of symmetry. The axis of symmetry has the equation </h2><h2>x=−b2a </h2><h2>The y-intercept of the equation is c. </h2><h2> When you want to graph a quadratic function you begin by making a table of values for some values of your function and then plot those values in a coordinate plane and draw a smooth curve through the points.</h2>

3 0

3 years ago