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2 years ago

Please please help I really need it today please show work

Mathematics

2 answers:

Cerrena [4.2K]2 years ago

4 0

Answer:

x=9.

perimeter=68

area=168

Step-by-step explanation:

First find the side that has a exact number which would be the bottom 28cm then look at the opposite side (2x+10) in order to find x subtract 10 from 28 which gives you 18, now 2x means x multiplied by 2 so what multiplied by 2 = 18? 9 meaning x=9. Now you are going to want to plug 9 in to your short side (x-3) once you plug it in it should look like this 9-3 and 9-3=6 meaning the short side is 6cm. In order for you to find the perimeter you have to add the sides together so 28+28+6+6 wich =68 and in order for you to find the area you have to multiply the length and the width so 28x6 which is 168.

hope that helped :)

kumpel [21]2 years ago

3 0

Answer:

x = 9 cm

Area = 168 square cm

Perimeter = 68 cm

Step-by-step explanation:

We know that opposite sides of a rectangle are equal.

$2x+10=28$

Subtract 10 from both sides.

$2x=28-10$

$2x=18$

Divide both sides by 2.

$x=9$

Therefore, the value of x is 9.

Length of the rectangle : l = 28 cm

Width of the rectangle : w = x-3 = 9-3 =6 cm

Area of rectangle is

$Area=l\times w=28\times 6=168\text{ cm}^2$

Perimeter of rectangle

$Perimeter=2(l+w)=2(28+6)=2(34)=68\text{ cm}$

Therefore, the area is 168 square cm and perimeter is 68 cm.

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Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

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Solution to the problem

Part a

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

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