Help on this pleasee

Tracy works at the Pizza Place. She earns $4.50 an hour plus tips. Last week, she earned a total

sweet [91]

Answer:

She work <u>23 hours</u>.

Step-by-step explanation:

Given:

Tracy works at the Pizza Place. She earns $4.50 an hour plus tips.

Last week, she earned a total of $156.00, of which $52.50 was tips.

Now, to find the number of hours she work.

Total salary = $156.00

Amount of tips = $52.50.

So, first we deduct the amount of tips from the total salary:

$\$156.00-\$52.50\\=\$103.50.$

<em>Thus, the remaining salary = $103.50.</em>

As, given salary she earns an hour = $4.50.

And then, to get the number of hours she work last week by dividing $103.50 the remaining salary by $4.50:

$103.50\div 4.50$

$=23\ hours.$

Therefore, she work 23 hours.

8 0

3 years ago

Find their total volume if x=10 feet, y=12 feet, and z= 7 feet

saveliy_v [14]

The volume would be 840 feet

4 0

3 years ago

Please help I will mark you as brainliest! Work out the area of shape P!

ozzi

To find the area of a triangle, multiply the height (6) by the base (4) and divide by 2.
- 6 • 4 = 24, 24 / 2 = 12.
The area of this triangle is 12.

7 0

3 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

What is the population standard deviation of this data set?*

Minchanka [31]

Answer:

C) 2

Step-by-step explanation:

step 1: Find mean of data set

2+4+4+5+7+8 = 30

30/6 = 5

Mean = 5

step 2: subtract each data value from the mean and square it

5-2 = 3; 3² = 9

5-4 = 1; 1² = 1

5-5 = 0; 0² = 0

5-7 = -2; (-2²) = 4

5-8 = -3; (-3²) = 9

Add the squared results:

9+1+1+0+4+9 = 24

Divide 24 by 6 to get the Variance of 4

Take the square root of the Variance to get the Standard Deviation

$\sqrt{4}$ = 2

5 0

3 years ago