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3 years ago

The length of a rectangle is (3x - 5) inches, and its width is 2x inches. Find the area of the rectangle.

Mathematics

2 answers:

n200080 [17]3 years ago

8 0

Answer:

6x^2 - 10x inches ^2

Step-by-step explanation:

The area of a rectangle is

A = l*w

= (3x-5) * (2x)

= 6x^2 - 10x

SVETLANKA909090 [29]3 years ago

7 0

Answer:

The area of the rectangle is 6x^2(squared)-10x

Step-by-step explanation:

To find the area of rectangles, you must multiply the length by the width. The formst for this is A=LxW. When you substitute each part of the equation in, it becomes 2x(3x-5). When you have parentheses, distribute the number on the outside to each number(or variable) on the inside. First, multiply 2x by 3x. This leaves you with 6x^2(squared). Then, multiply the 2x by the -5. This leaves you with -10x. Combine each part to get 6x^2(squared)-10x. :)

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It is claimed that automobiles are driven on average more than 20,000 kilometers per year. To test this claim, 100 randomly sele

kow [346]

Answer:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.

Step-by-step explanation:

1) Data given and notation

$\bar X=23500$ represent the sample mean

$s=3900$ represent the sample standard deviation

$n=100$ sample size

$\mu_o =2000$ represent the value that we want to test

$\alpha$ 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 (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mean is higher than 20000, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2000$

Alternative hypothesis: $\mu > 2000$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=100-1=99$