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3 years ago

15

Katie plans to paint the entire surface of the box. What is the surface area of the prism that Katie will paint? (Note: 12 inche

s = 1 foot)

1 answer:

Zina [86]3 years ago

4 0

Answer: A = 1032 in²

Step-by-step explanation:

two sides each of rectangles

2(10)(18) + 2(10)(12) + 2(12)(18) = 1032 in²

You might be interested in

The following dotplot shows the centuries during which the 11 castles whose ruins remain in Somerset, England were constructed.

nasty-shy [4]

Taking into account the rule 1.5 IQR, there aren't high outliers in the displayed data set.

Calculation of the Interquartile Range and the rule 1.5 IQR.

To calculate the Interquartile Range IQR, you must subtract the value of the first quartile Q1, to the third quartile Q3, as you can see in the formula below:

IQR = Q3 - Q1

Using the five-number summary, you can replace the formula:

IQR = 17 - 13

IQR = 4

Before to apply the rule 1.5 IQR, you must multiply the IQR by 1.5:

4 * 1.5 = 6

Now, to apply the rule you must add the value obtained (6), to the third quartile (17), if any data is over this, theoretically it's an outlier:

17 + 6 = 23 (As you do not have any data over this value, it is identified that there are no outliers in the upper section).

Finally, you must subtract the value obtained (6) to the first quartile (13), if any data is under this, theoretically it's an outlier:

13 - 6 = 7 (Since you don't have any castle data built before this century, there wouldn't be any theoretically outliers either).

By this reason, using the rule of 1.5 IQR, you can see there aren't outliers in the displayed data set.

brainliest please if correct.

3 0

2 years ago

Let H and K be subgroups of a group G, and let g be an element of G. The set <img src="https://tex.z-dn.net/?f=%5Cmath%20HgK%20%

34kurt

Answer:

Yes, double cosets partition G.

Step-by-step explanation:

We are going to define a <em>relation</em> over the elements of G.

Let $x,y\in G$ . We say that $x\sim y$ if, and only if, $y\in HxK$ , or, equivalently, if $y=hxk$ , for some $h\in H, k\in K$ .

This defines an <em>equivalence relation over </em><em>G</em>, that is, this relation is <em>reflexive, symmetric and transitive:</em>

Reflexivity: ( $x\sim x$ for all $x\in G$ .) Note that we can write $x=exe$ , where $e$ is the <em>identity element</em>, so $e\in H,K$ and then $x\in HxK$ . Therefore, $x\sim x$ .
Symmetry: (If $x\sim y$ then $y\sim x$ .) If $x\sim y$ then $y=hxk$ for some $h\in H$ and $k\in K$ . Multiplying by the inverses of h and k we get that $x=h^{-1}yk^{-1}$ and is known that $h^{-1}\in H$ and $k^{-1}\in K$ . This means that $x\in HyK$ or, equivalently, $y\sim x$ .

Transitivity: (If $x\sim y$ and $y\sim z$ , then $x\sim z$ .) If $x\sim y$ and $y\sim z$ , then there exists $h_1,h_2\in H$ and $k_1,k_2\in K$ such that $y=h_1xk_1$ and $z=h_2yk_2$ . Then, $\\ z=h_2yk_2=h_2(h_1xk_1)k_2=(h_2h_1)x(k_1k_2)=h_3xk_3$ where $h_3=h_2h_1\in H$ and $k_3=k_1k_2\in K$ . Consequently, $z\sim x$ .

Now that we prove that the relation " $\sim$ " is an equivalence over G, we use the fact that the <em>different equivalence classes partition </em><em>G.</em><em> </em>Since the equivalence classes are defined by $[x]=\{y\in G\colon x\sim y\} =\{y\in G \colon y=hgk\ \text{for some } h\in H, k\in K \}=HxK$ , then we're done.

5 0

3 years ago

If the current time is 3:15 am right now what time will it be in 3 hours in 20 mins from now

poizon [28]

6:35 a.m.

1) To find out what time it is, then let's add them up:

hours minutes

3 15

3 20

--------------------------

6 35

2) Note that since the sum of the minutes does not pass 60 minutes, then we can end up right here.

3) Hence, the answer is 6:35 a.m.

5 0

1 year ago

50% of cakes baked are party cakes - 1/5 were fruit cakes and remainder were sponge cakes. What percentage of cakes were sponge

Katen [24]

Start by think that 50% can be looked at as 5/10.
Out of the remaining 5/10, 1/5 is fruit cakes, leaving the remaining 4/5 to sponge cakes. Which means that 40% of the cakes were sponge cakes.

50% = party cakes
10% = fruit cakes
40% = sponge cakes

3 0

3 years ago

Read 2 more answers

For the following relation, complete the table of values and sketch the graph.

Gekata [30.6K]

The table of values for the given equation is presented below:

x y

-3 17

-2 2

-1 -7

0 -10

1 -7

2 2

3 17

<h3>Quadratic Function</h3>

The Standard form for a quadratic equation is ax²+ bx + c=0, where: a, b and c are your respective coefficients. In the quadratic function, the coefficient "a" must be different than zero (a≠0) and the degree of the function must be equal to 2.

To solve a quadratic function, you should find the discriminant: D=b²-4ac and then use this variable in the formula: $x=\frac{-b \±\sqrt{\Delta} }{2a}$ .

The given equation is a quadratic function because have a degree equal to 2. Then, when you plot the graph, you obtain a parabola.

First, you should replace the given values for x in the equation y=3x²-10

x y

-3 y=3*(-3)²-10= 3*9-10=27-10=17

-2 y=3*(-2)²-10= 3*4-10=12-10=2

-1 y=3*(-1)²-10= 3*1-10=3-10= -7

0 y=3*(0)²-10= 3*0-10=-10= -10

1 y=3*(1)²-10= 3*1-10=3-10= -7

2 y=3*(2)²-10= 3*4-10=12-10= 2

3 y=3*(3)²-10= 3*3-10=27-10= 17

Now, you have the points to draw the graph, show the attached image.

Read more about the quadratic function here:

brainly.com/question/1497716#

#SPJ1

4 0

1 year ago