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3 years ago

Find the quotient: 15x^4y^2 - 25x^2y^3 + 45xy / 5xy:

Mathematics

1 answer:

salantis [7]3 years ago

6 0

Answer:

3x^3y − 5xy^2 + 9

Step-by-step explanation:

When there is only one denominator term, the quotient can be found by dividing the numerator terms individually.

$\dfrac{15x^4y^2 - 25x^2y^3 + 45xy}{5xy}=\dfrac{15x^4y^2}{5xy}-\dfrac{25x^2y^3}{5xy}+\dfrac{45xy}{5xy}=3x^3y-5xy^2+9$

_____

There are a couple of things to keep in mind here:

an exponent tells the number of times the factor is repeated
equal numbers of identical factors cancel from numerator and denominator, for example x^3/x = (x·x·x)/x = x·x = x^2. (One factor of x cancels.)

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Find the area of a circle with radius, r = 76cm.<br> Give your answer rounded to 3 SF.

PSYCHO15rus [73]

Answer:

17,700.00

Step-by-step explanation:

Area of a circle = 22

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=22

------ 76x76 =123,750

7 ---------- =1,768.571

1,768.571 rounded to 3 SF=17700.00

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2 years ago

What the length of a diagonal of a square with perimeter of 16

Vladimir79 [104]

A- the length of the side of a square
P = 16

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therefore d = 4√2

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3 years ago

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schepotkina [342]

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3 years ago

Let x be a Poisson random variable with μ = 9.5. Find the probabilities for x using the Poisson formula. (Round your answers to

miv72 [106K]

Answer: 1

Step-by-step explanation: the probability mass function that defines a possion probability distribution is given below as

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For this question, x = 0 and u = 9.5

Hence we have that

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3 years ago

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average re

Zolol [24]

Answer:

a) $df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

b) $df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

c) $df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X$ represent the sample mean

$s$ represent the standard deviation for the sample

$n$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:

Null hypothesis: $\mu \leq 20$

Alternative hypothesis: $\mu > 20$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

3 0

3 years ago