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4 years ago

What sales tax do you pay if you buy a computer for $229 and the tax rate is 5%

Mathematics

1 answer:

kozerog [31]4 years ago

5 0

The answer is
$11.45

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How is mode different from the median?

ycow [4]

The median is the middle number, while the mode is the number that repeats the most.

Hope this helps!

5 0

3 years ago

Jeremiah has to spend $40 to buy candy. He then sold candy for $1.50 per piece. If Jeremiah was able to sell 60 pieces of the ca

Nadya [2.5K]

Answer:

He made $50

Step-by-step explanation:

1.5 times 60 equals 90, but seeing how he spent $40 to buy the candy he only ends up with $50 in profits

7 0

3 years ago

Identify the 7th term of the geometric sequence in which a2 = 324 and a4 = 36.

ziro4ka [17]

Answer:

a7=4/3 or a7=-4/3

Step-by-step explanation:

using the geometric sequences formula

an=ar^n-1

a2=ar

a4=ar³

when a2=324 and a4=36

324=ar...........(1)

36=ar³............(2)

from equation (1) a=324/r substitute in equation (2)

we have :

36=324/r *r³

36=324r²

r²=36/324

r²=1/9

r=±1/3

substitute when r=±1/3 in (1)

324=a(±1/3)

a=±972

so the 7th term is

when r=±1/3

we have

a7=ar^6

a7=±972(±1/3)^6

a7=972/729

a7=4/3 or a7=-4/3

3 0

2 years ago

Plz hury plz What type of graph is best to use to organize numerical data based on their digits to compare value?

katovenus [111]

Answer:

Step-by-step explanation:

6 0

3 years ago

A force of 8 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it

-BARSIC- [3]

Answer:

The work done in stretching it from its natural length to 14 in. beyond its natural length is W=8.17 ft-lb.

Step-by-step explanation:

We know that a force of 8 lb is required to hold a spring stretched 8 in. beyond its natural length.

This let us calculate the spring constant k as:

$F=kx\\\\k=F/x=(8 \;lb)/(8\;in)=1\;lb/in$

We know that work is, in an scalar form, the product of force and distance.

The force F is equal to the spring constant multiplied by the distance from the natural length.

Then, as the force changes with the distance from the natural length, we have to calculate integrating:

$W=\int_0^{14}F\,dx=\int_0^{14}kx\,dx\\\\\\W=\dfrac{1}{2}kx^2\left|^{14}_0=\dfrac{1}{2}(1\,lb/in)(14 \,in)^2-\dfrac{1}{2}(1\,lb/in)(0 \,in)^2$

$W=98lb\cdot in-0lb-in\\\\W=98(lb\cdot in)\cdot \dfrac{1 \,ft}{12\,in}=8.17\;lb\cdot ft$

5 0

3 years ago