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3 years ago

The area of a rectangle is 0.8 square units. The length is 3.2 units and the width is x units.

Mathematics

2 answers:

KengaRu [80]3 years ago

6 0

Answer:

Option 1 - x=0.25

Step-by-step explanation:

Given : The area of a rectangle is 0.8 square units. The length is 3.2 units and the width is x units.

To find : What is the value of x?

Solution :

The area of the rectangle is given by,

$A=L\times W$

The length of the rectangle is L=3.2 units

The width of the rectangle is W=x units

The area of the rectangle is A=0.8 square units

Substitute the value in the formula,

$0.8=3.2\times x$

$x=\frac{0.8}{3.2}$

$x=0.25$

Therefore, Option 1 is correct.

The value of x is 0.25.

Naddika [18.5K]3 years ago

3 0

Area= l*b
0.8=3.2*x
0.8/3.2=x
x=0.25 units

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A simple random sample of 28 Lego sets is obtained and the number of pieces in each set was counted.The sample has a standard de

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Answer:

Step-by-step explanation:

Given that:

A simple random sample n = 28

sample standard deviation S = 12.65

standard deviation $\sigma$ = 11.53

Level of significance ∝ = 0.05

The objective is to test the claim that the number of pieces in a set has a standard deviation different from 11.53.

The null hypothesis and the alternative hypothesis can be computed as follows:

Null hypothesis:

$H_0: \sigma^2 = \sigma_0^2$

Alternative hypothesis:

$H_1: \sigma^2 \neq \sigma_0^2$

The test statistics can be determined by using the following formula in order to test if the claim is statistically significant or not.

$X_0^2 = \dfrac{(n-1)S^2}{\sigma_0^2}$

$X_0^2 = \dfrac{(28-1)(12.65)^2}{(11.53)^2}$

$X_0^2 = \dfrac{(27)(160.0225)}{132.9409}$

$X_0^2 = \dfrac{4320.6075}{132.9409}$

$X_0^2 = 32.5002125$

$X^2_{1- \alpha/2 , df} = X^2_{1- 0.05/2 , n-1}$

$X^2_{1- \alpha/2 , df} = X^2_{1- 0.025 , 28-1}$

From the chi-square probabilities table at 0.975 and degree of freedom 27;

$X^2_{0.975 , 27}$ = 14.573

$X^2_{\alpha/2 , df} = X^2_{ 0.05/2 , n-1}$

$X^2_{\alpha/2 , df} = X^2_{0.025 , 28-1}$

From the chi-square probabilities table at 0.975 and degree of freedom 27;

$X^2_{0.025 , 27}=$ 43.195

Decision Rule: To reject the null hypothesis if $X^2_0 \ > \ X^2_{\alpha/2 , df} \ \ \ or \ \ \ X^2_0 \ < \ X^2_{1- \alpha/2 , df}$ ; otherwise , do not reject the null hypothesis:

The rejection region is $X^2_0 \ > 43.195 \ \ \ or \ \ \ X^2_0 \ < \ 14.573$

Conclusion:

We fail to reject the null hypothesis since test statistic value 32.5002125 lies between 14.573 and 43.195.

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3 years ago