The length of a rectangle is 5 yd more than twice the width x. The area is 900 yd?.

Each of 11 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating n

Pie

Answer:

The correct answer is 1,51%

Step-by-step explanation:

To get the probability equally likely of the first 6 examined that have a defective compressor, we use the following formula

P=# of possibilities that meet the condition / #of equally likely possibilities.

As there are multiple events , and all must be achieve, we multiply the probabilities.

P1=7 / 11

P2=6 / 10

P3=5 / 9

P4=4 / 8

P5=3 / 7

P6=2 / 6

P(x)=7 / 11*6 / 10*5 / 9*4 / 8*3 / 7*2 / 6=0,01515151515

P(x)=1,51%

3 0

3 years ago

10 times The sum of half a numberand six is eight

GuDViN [60]

The numerical form of this worded expression is:

10(1/2n + 6)=8

4 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

An above ground swimming pool is leaking. After 1/2 hour the pool has leaked 7/8 of a gallon of water. How many gallons of water

Lilit [14]

1 $\frac{3}{4}$ , or 1.75<u />

6 0

3 years ago

Write an equation of a parabola with vertex at the origin and the given focus<br> (6,0)

Masteriza [31]

Answer:

Step-by-step explanation:

eq. of a parabola with vertex at origin and focus (a,0) is

y²=4ax

a=6

so y²=4×6×x

or y²=24 x

4 0

2 years ago