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3 years ago

Which histograms are approximately symmetric and bell shaped?

Mathematics

1 answer:

tankabanditka [31]3 years ago

3 0

Answer:

B and E

Step-by-step explanation:

im pretty sure what you have is correct

You might be interested in

Solve g(-1)=x^2-5x+2

olga55 [171]

Answer:

x = 5/2 + sqrt(17)/2 or x = 5/2 - sqrt(17)/2

Step-by-step explanation:

Solve for x:

x^2 - 5 x + 2 = 0

Subtract 2 from both sides:

x^2 - 5 x = -2

Add 25/4 to both sides:

x^2 - 5 x + 25/4 = 17/4

Write the left hand side as a square:

(x - 5/2)^2 = 17/4

Take the square root of both sides:

x - 5/2 = sqrt(17)/2 or x - 5/2 = -sqrt(17)/2

Add 5/2 to both sides:

x = 5/2 + sqrt(17)/2 or x - 5/2 = -sqrt(17)/2

Add 5/2 to both sides:

Answer: x = 5/2 + sqrt(17)/2 or x = 5/2 - sqrt(17)/2

5 0

3 years ago

Your new job is at the custom t shop, where t-shirts are printed to order. For each order, custom t shop charges $8.00 per shirt

Rasek [7]

Answer:

Price = 15+8t

Step-by-step explanation:

Your new job is at the custom t shop, where t-shirts are printed to order.

For each order, custom t shop charges $8.00 per shirts plus a one time set up fee for $15.00.

Here, $15 remain fix and $8 increases with number of shirts.

Price = 15+8t

Hence, the price of the t-shirts is "15+8t".

6 0

3 years ago

Solve for n: 0.22n = 6.6. <br><br><br>

vagabundo [1.1K]

Answer:

Step-by-step explanation:

Solve for n: 0.22n = 6.6.

6.6/0.22 (because for example 3/3 and n/n= 1 so 0.22/0.22 = 1 so we have to divide 6.6 since it is part of the equation and we have to do it on both sides.)
=30

3 0

3 years ago

A quadrilateral has vertices A, B, C, and D. A line of reflection is drawn so that A is 6 units away from the line, B is 4 units

Kryger [21]

Answer:

The correct option is O B'

Step-by-step explanation:

We have a quadrilateral with vertices A, B, C and D. A line of reflection is drawn so that A is 6 units away from the line, B is 4 units away from the line, C is 7 units away from the line and D is 9 units away from the line.

Now we perform the reflection and we obtain a new quadrilateral A'B'C'D'.

In order to apply the reflection to the original quadrilateral ABCD, we perform the reflection to all of its points, particularly to its vertices.

Wherever we have a point X and a line of reflection L and we perform the reflection, the new point X' will keep its original distance from the line of reflection (this is an important concept in order to understand the exercise).

I will attach a drawing with an example.

Finally, we only have to look at the vertices and its original distances to answer the question.

The vertice that is closest to the line of reflection is B (the distance is 4 units). We answer O B'

8 0

3 years ago

Let Y be a random variable with a density function given by

Neporo4naja [7]

From the given density function we find the distribution function,

$F_Y(y)=P(Y\le y)=\displaystyle\int_{-\infty}^y f_Y(t)\,\mathrm dt=\begin{cases}0&\text{for }y$

(a)

$F_{U_1}(u_1)=P(U_1\le u_1)=P(3Y\le u_1)=P\left(Y\le\dfrac{u_1}3\right)=F_Y\left(\dfrac{u_1}3\right)$

$\implies F_{U_1}(u_1)=\begin{cases}0&\text{for }u_1$

$\implies f_{U_1}(u_1)=\begin{cases}\frac{{u_1}^2}{18}&\text{for }-3\le u_1\le3\\0&\text{otherwise}\end{cases}$

(b)

$F_{U_2}(u_2)=P(3-Y\le u_2)=P(Y\ge3-u_2)=1-P(Y$

$\implies F_{U_2}(u_2)=\begin{cases}0&\text{for }u_2$

$\implies f_{U_2}(u_2)=\begin{cases}\frac32(u_2-3)^2&\text{for }2\le u_2\le4\\0&\text{otherwise}\end{cases}$

(c)

$F_{U_3}(u_3)=P(Y^2\le u_3)=P(-\sqrt{u_3}\le Y\le\sqrt{u_3})=F_Y(\sqrt{u_3})-F_y(\sqrt{u_3})$

$\implies F_{U_3}(u_3)=\begin{cases}0&\text{for }u_3$

$\implies f_{U_3}(u_3}=\begin{cases}\frac32\sqrt u&\text{for }0\le u\le1\\0&\text{otherwise}\end{cases}$

5 0

3 years ago