Ask question

Ask question

All categories

3 years ago

15

Samantha deposit 300 in an account that earns an annual interest rate of 2.5 after nine months she computes the simple interest.

What mistakes did Samantha make?

1 answer:

antiseptic1488 [7]3 years ago

4 0

Answer:

She is doing a mistake of calculating interest after 9 months in place of after 12 months.

Step-by-step explanation:

Samantha deposit $300 in an account that earns an annual interest rate of 2.5%.

Now, Samantha after nine months of deposit computes the simple interest.

She is doing a mistake of calculating interest after 9 months in place of after 12 months.

The calculation of interest should be on a yearly basis (i.e. 12 months) as the interest rate is 2.5% per year. (Answer)

You might be interested in

sharon earns $25 per item that she sells plus a base salary of $150 per week. write and solve an inequality to find how many ite

lozanna [386]

She has to sell 24 items

7 0

3 years ago

a basketball player through the ball vertically in the air. the height h of the ball after time t is given by the quadratic func

Flauer [41]

I think yes he did in ball rolling in throw the ball 5_t+7t. yes maybe in be ball player in won

4 0

3 years ago

Identify the solution set for: −2|−2x−4|=−12

myrzilka [38]

Answer:

C: {−5, 1)

Step-by-step explanation:

Starting with −2|−2x−4|=−12, divide both sides by -2:

|−2x−4| = 6

Reduce this by dividing all terms by 2:

|-x - 2| = 3

This is equivalent to x + 2 = 3 and x + 2 = - 3

Solving thse two equations, we get x = 1 and x = -5

And so the solution set is C: {−5, 1)

4 0

3 years ago

Read 2 more answers

Find an equation of the sphere with center (−2, 3, 7) and radius 7.What is the intersection of this sphere with the yz-plane?

masya89 [10]

Answer:

The equation of the sphere with center (-2, 3, 7) and radius 7 is $(x+2)^{2}+(y-3)^{2}+(z-7)^{2} = 49$ .

The intersection of the sphere with the yz-plane is $(y-3)^{2}+(z-7)^{2} = 49$ .

Step-by-step explanation:

We know that any sphere can be represented by the following equation:

$(x-h)^{2}+(y-k)^{2}+(z-s)^{2} = r^{2}$

Where:

$h$ , $k$ , $s$ - Coordinates of the center of the sphere, dimensionless.

$r$ - Radius of the sphere, dimensionless.

If $(h,k, s) = (-2,3,7)$ and $r = 7$ , we obtain this expression:

$(x+2)^{2}+(y-3)^{2}+(z-7)^{2} = 49$

The intersection of the sphere with the yz-plane observe the following conditions:

$x = 0$ , $y \in \mathbb {R}$ , $z\in \mathbb{R}$

Hence, the expression above can be reduced into this:

$(y-3)^{2}+(z-7)^{2} = 49$

6 0

3 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