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MrRa [10]
3 years ago
5

Evaluate 4k2 + 3 when k = 5

Mathematics
2 answers:
ElenaW [278]3 years ago
8 0

\text{Hey there!}

\text{If k = 5 the replace the 'k' value with 5}

\text{4(5)}^2+3

\text{(5)}^2=5\times5=25

\text{4 (25)+ 3 = ?}

\text{4 (25) = 25 + 25 + 25 + 25 = 100}

\text{100 + 3 = 103}

\boxed{\boxed{\bf{Your\ answer: 103}}}\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

lyudmila [28]3 years ago
7 0

Answer:

103

Step-by-step explanation:

Evaluate 4k2 + 3

To evaluate 4k² + 3, when k = 5, which means when you sees k, put 5 in replacement

4k² + 3 = 4(5)² + 3

4(5×5) + 3

4(25) + 3

open the bracket

4 × 25 = 100

∴ 100 + 3 = 103

Please mark me brainliest  

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Answer:

There is not enough evidence to support the claim that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

P-value=0.042.

Step-by-step explanation:

The question is incomplete:

The data of the scores for each student is:

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430        465

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We will generate a sample for the difference of scores (before - after) and test that sample.

The sample of the difference is [35 -10 15 50 -15 5 25 10 5 0 10 5 -10]

This sample, of size n=13, has a mean of 9.615 and a standard deviation of 18.423.

The claim is that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

Then, the null and alternative hypothesis are:

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The significance level is 0.01.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{18.423}{\sqrt{13}}=5.11

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{9.615-0}{5.11}=\dfrac{9.615}{5.11}=1.882

The degrees of freedom for this sample size are:

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This test is a right-tailed test, with 12 degrees of freedom and t=1.882, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>1.882)=0.042

As the P-value (0.042) is bigger than the significance level (0.01), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

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nasty-shy [4]

When we Simplify [(x^2)^3 × 5x] / [6x^2 × 15x^3], the result obtained is (1/18)x^2

<h3>Data obtained from the question</h3>
  • [(x^2)^3 × 5x] / [6x^2 × 15x^3]
  • Simplification =?

<h3>How to simplify [(x^2)^3 × 5x] / [6x^2 × 15x^3]</h3>

[(x^2)^3 × 5x] / [6x^2 × 15x^3]

Recall

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Thus,

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Recall

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Thus,

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Recall

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Thus,

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Range is largest number minus smallest number

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