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2 years ago

5

Write an expression for the area of square 4 by combining the areas of the four triangles and the two squares.

1 answer:

lesya [120]2 years ago

6 0

The area of Square 4 is a2+2ab+b2

The area of Square 5 is 2ab+c2

Setting them equal to one another (a2+2ab+b2=2ab+c2) and simplifying yields that a2+b2=c2 (note that 2ab is what they have in common), which is Pythagoras' theorem.

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Distribute and simplify these radicals. 23-(V+ 3) O 2V5 + 6 02/15 O 30 O 6/2+6

zhannawk [14.2K]

Answer:

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Step-by-step explanation:

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3 years ago

Katie jogged a total distance of 8 and 1 over 2 miles during the months of January and February. If Katie only jogged 1 over 4 m

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For each of the following, decide whether the description fits the SD or the SE.

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3 years ago

What is four and 1 fith divided by 2 thirds

Sloan [31]

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A researcher tests five individuals who have seen paid political ads about a particular issue. These individuals take a multiple

devlian [24]

Answer:

$t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606$

The degrees of freedom are given by:

$df=n-1=5-1=4$

The p value wuld be given by:

$p_v =2*P(t_{(4)}>2.606)=0.060$

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Step-by-step explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=46$ represent the mean height for the sample

$s=5.148$ represent the sample standard deviation

$n=5$ sample size

$\mu_o =40$ represent the value that we want to test

$\alpha=0.05$ 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

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:

Null hypothesis: $\mu = 40$

Alternative hypothesis: $\mu \neq 40$

The statistic is given by:

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

Replacing we got:

$t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606$

The degrees of freedom are given by:

$df=n-1=5-1=4$

The p value wuld be given by:

$p_v =2*P(t_{(4)}>2.606)=0.060$

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

6 0

4 years ago