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Stells [14]
1 year ago
15

A rational expression simplifies to 3. the denominator of the original expression is given. which polynomial is the numerator?

Mathematics
1 answer:
4vir4ik [10]1 year ago
4 0

The polynomial within the numerator is 9x²+45x+54.

Given the denominator of the initial expression is \frac{?}{3x^2+15x+18} and also the rational expression simplifies to three.

A mathematical expression which will be rewritten to a rational fraction, an algebraic fraction specified both the numerator and therefore the denominator are polynomials. and a polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

Let's take the polynomial within the numerator is k.

So, rewrite the given expression as

\frac{k}{3x^2+15x+18}=3

or it also can be written as

\frac{k}{3x^2+15x+18}=\frac{3}{1}

Cross multiply either side as a×d=b×c where a=k, b=3x²+15x+18, c=3 and d=1 and acquire

k=3(3x²+15x+18)

Simplify the above expression,

k=9x²+45x+54

Hence, the polynomial within the numerator within the given expression \frac{?}{3x^2+15x+18} which is simplified to 3 is 9x²+45x+54.

Learn about rational expression from here brainly.com/question/3764404

#SPJ4

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Step-by-step explanation:

the circumference of a circle is

C = 2×pi×r

with r being the radius.

the square side length is 22 cm. so, the circumference or perimeter of the square (and therefore the length of the wire) is

4×22 = 88 cm.

now, it is the same wire of the same length that is now forming a circle.

so, the circumference of the square is also the circumference of the circle.

therefore,

88 = 2×pi×r

44 = pi×r

r = 44/pi = 14.00563499... cm

so, rounded, radius = 14 cm.

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Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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