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Kay [80]
2 years ago
12

Which is the simplified form of the expression (StartFraction (x Superscript negative 3 Baseline) (y squared) Over (x Superscrip

t 4 Baseline) (y superscript 6 Baseline) EndFraction) cubed?
Mathematics
2 answers:
Vlad [161]2 years ago
8 0

Answer:

The simplified form of the expression is \frac{1}{x^{7} y^{4}}.

Step-by-step explanation:

The expression is:

\frac{x^{-3}\cdot y^{2}}{x^{4}\cdot y^{6}}

The division rule of exponents is:

a^{m}\div a^{n}=a^{m-n}

The negative exponent rule is:

a^{-m}=\frac{1}{a^{m}}

Simplify the expression as follows:

\frac{x^{-3}\cdot y^{2}}{x^{4}\cdot y^{6}}=[x^{-3-4}]\times [y^{2-6}]

         =x^{-7}\times y^{-4}\\\\=\frac{1}{x^{7} y^{4}}

Thus, the simplified form of the expression is \frac{1}{x^{7} y^{4}}.

Ipatiy [6.2K]2 years ago
6 0

B) 1/x^21 y^12

I just took the quiz on edge

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On a piece of paper, graph this system of inequalities. Then determine which region contains the solution to the system.
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B or C  hope its helps...

 

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Which of the following slopes show that the set of points C(1, 1), D(3, -4), E(5, 8) are not collinear?
MArishka [77]
C is the answer you want

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A rope is 66 feet long. It is cut into two pieces such that one piece is half the length of the other. What is the length of the
dedylja [7]
So if one bit of the length, is half the size of the other bit then we can make the following equation, for x being the length of rope:

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2 years ago
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Pierre lifts a 60-kg crate onto a truck bed 1 m high in 3 s. Marie lifts twenty-four 2.5-kg boxes onto the same truck bed in a t
lys-0071 [83]

Answer:

Pierre operates at a greater power level than Marie.

Marie and Pierre do  the same amount of work

Step-by-step explanation:

Mass of crate=60kg

Height of truck bed=1 m

Time=3s

Pierre's work done=60\times 1\times 9.8=588J

Using formula Work done=mgh

Where m=mass of object

g=Acceleration due to gravity=9.8m/s^2

Power=\frac{w}{t}

Using the formula

Pierre's power=\frac{588}{3}=196watt

Mass of box=2.5Kg

Total number of boxes=24

Total mass of boxes=2.5\times 24=60kg

Time=2 min=2\times 60=120s

1 min=60 s

Marie;s work done=60\times 1\times 9.8=588J

Marie's power=\frac{588}{120}=4.9watt

Pierre operates at a greater power level than Marie.

Marie and Pierre do  the same amount of work

7 0
2 years ago
A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average
larisa [96]

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: \mu \leq  94

Alternate Hypothesis: \mu > 94

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}

Using the values, we get:

t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

\alpha =0.05

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

3 0
2 years ago
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