All categories

2 years ago

1. Jayron worked for 7.5 hours and got paid $63.75. How much does Jayron

Mathematics

2 answers:

Kazeer [188]2 years ago

6 0

The correct answer : $8.50 / 1 hour

denpristay [2]2 years ago

6 0

Answer:

$8.50/ 1 hour

Step By Step Explanation:

To find how much someone makes per hour, you have to divide how much money they made by how long it took them to make that money.

Therefore,

If Jayron makes $63.75 after working for 7.5 hours, to find how much he makes per hour you would have to divide 63.75 by 7.5, which is 8.5.

You might be interested in

Which quadrilateral does NOT have two pairs of parallel sides?

WINSTONCH [101]

It a................

8 0

2 years ago

Sales price is 67.50 sales tax is 5% how much is the total amount paid

Zinaida [17]

$70.88

x/67.5=5/100
cross multiply
100x=337.5
divide 100
x=3.375 or 3.38

67.50+3.38= 70.88

7 0

3 years ago

How many time larger is 5000 to 5

disa [49]

Answer:

1000 times larger.

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

2 years ago

A tissue box has a height of 4 inches, a width of 8 inches and a length of 5.5 inches. What is the volume in inches cubed?

denis-greek [22]

Answer:

Your answer is going to be 176

7 0

2 years ago