Please help with this math question

Tina is packing for a five-day trip. She has one blue, one red, one white, one green, and one grey t-shirt. Tina will not wear e

agasfer [191]

Answer:

This means that Tina has enough t-shirts for the entire trip and doesn't need to pack any more t-shirts.

Step-by-step explanation:

In total, she has five t-shirts, and she is packing for a five-day trip. Therefore, she has enough shirts to last her the entire trip.

5 0

2 years ago

Compute the probability of random sly selecting a queen or diamond.

FrozenT [24]

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3 0

3 years ago

The cost C (in dollars) to participate in a ski club is given by the literal equation C=85x+60, where X=C/85-12/17 is the number

dybincka [34]

Given:

The cost C (in dollars) to participate in a ski club is given by,

$\begin{gathered} C=85x+60 \\ (or) \\ x=\frac{C}{85}-\frac{12}{17} \end{gathered}$

Where x is the number of ski trips we take.

To find:

The number of ski trips when a total cost of $315 and $485 is spent.

Explanation:

Substituting C = 315 in the given equation we get,

$\begin{gathered} x=\frac{315}{85}-\frac{12}{17} \\ x=\frac{315-60}{85} \\ x=\frac{255}{85} \\ x=3\text{ ski trips} \end{gathered}$

Substituting C = 485 in the given equation we get,

$\begin{gathered} x=\frac{485}{85}-\frac{12}{17} \\ x=\frac{485-60}{85} \\ x=\frac{425}{85} \\ x=5\text{ ski trips} \end{gathered}$

Thus,

If we spend a total cost of $315, we can take 3 ski trips.

If we spend a total cost of $485, we can take 5 ski trips.

Final answer:

• If we spend a total cost of $315, we can take 3 ski trips.

,

• If we spend a total cost of $485, we can take 5 ski trips.

8 0

1 year ago

Write the following ratio in their simplest form (a)150minutes:1 1/2 hours (b) 150cm:1060mm

spayn [35]

Answer:

a) 5:3

b) 7.5:5.3

Step-by-step explanation:

150min : 90min (1 and a half hour)

10 : 6

5 : 3

150 × 10 mm : 1060 mm

1500 : 1060

750 : 530

7.5 : 5.3

7 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 relation 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 equivalence relation over G, that is, this relation is reflexive, symmetric and transitive:

Reflexivity: ( $x\sim x$ for all $x\in G$ .) Note that we can write $x=exe$ , where $e$ is the identity element, 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 different equivalence classes partition G. 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