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alexgriva [62]
4 years ago
11

Which expressions are equivalent to 25x^4–64? select three options.

Mathematics
2 answers:
vichka [17]4 years ago
8 0

Answer: it’s A, B,C

Step-by-step explanation:

BARSIC [14]4 years ago
6 0
<h2>The equivalent expression of 25x^4 - 64 are "a), b) and c)".</h2>

Step-by-step explanation:

The given expression is:

25x^4 - 64

To find, the equivalent expressions are:

a) 25x^4 + 40x - 40x - 64

=  25x^4 - 64, is the equivalent expression.

b) 25x^4 + 13x - 13x - 64

=  25x^4 - 64, is the equivalent expression.

c) (5x^2 + 8)(5x^2 - 8)

Using the algebraic identity,

(a + b)(a - b) = a^{2}-b^{2}

= (5x^2)^2-8^2

=  25x^4 - 64, is the equivalent expression.

d) (x^2 + 13)(x^2 - 13)

Using the algebraic identity,

(a + b)(a - b) = a^{2}-b^{2}

= (x^2)^2-13^2

=  x^4 - 169, is not a equivalent expression.

e) (5x^2-8)^2

Using the algebraic identity,

(a-b)^{2} =a^{2} -2ab+b^2

= (5x^2)^{2} -2(5x^2)(8)+8^2

= 25x^4 -80x^2+64,  is not a equivalent expression.

∴ The equivalent expression of 25x^4 - 64 are "a), b) and c)".

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It has been hypothesized that the standard deviation of the germination time of radish seeds is 8 days. The standard deviation o
Yanka [14]

Answer:

\chi^2 =\frac{60-1}{64} 36 =33.19

The degrees of freedom are:

df= n-1= 60-1= 59

And the p value would be given by:

p_v =2*P(\chi^2 >33.19)=0.0053

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true deviation is different from 8 days.

Step-by-step explanation:

Data given

n=60 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =6^2 =36 represent the sample variance obtained

\sigma^2_0 =8^=64 represent the value that we want to test

Hypothesis to test

We want to verify if the true deviation is equal to 8 days o not, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 = 64

Alternative hypothesis: \sigma^2 \neq 64

The statistic is given by:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

Replacing we got:

\chi^2 =\frac{60-1}{64} 36 =33.19

The degrees of freedom are:

df= n-1= 60-1= 59

And the p value would be given by:

p_v =2*P(\chi^2 >33.19)=0.0053

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true deviation is different from 8 days.

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4 years ago
( 16 points ) Fill in the blanks 11 - = 5
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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5) correct

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Step-by-step explanation:

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