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3 years ago

Which one of the following formulas defines the area, A A , of a square in terms of its perimeter, p ?

1 answer:

6 0

A square has 4 sides of equa length

the perimiter is 4 times the side, or P=4s where s=length of 1 side

the area is length times witdth, but since the length=width, area=s² where s is the length of one side

so express A=s² in terms of P when P=4s

solve for s first in P=4s
P=4s
divide both sides by 4
$\frac{P}{4}=s$

subsitute $\frac{P}{4}$ for s in the other eqaution
$A=s^2$
$A=(\frac{P}{4})^2$
$A=\frac{P^2}{16}$

in terms of the perimiter, the area is $A=\frac{P^2}{16}$

You might be interested in

What’s the midpoint P(0,-2), Q(1,1

LiRa [457]

You would have to add x1 + x2 and divide by 2. do the same with y1 and y2.

(1/2,-1/2)

6 0

3 years ago

An auto company claims that the fuel efficiency of its sedan has been substantially improved. A consumer advocate organization w

zzz [600]

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021$

$df=n-1=12-1=11$

$p_v =P(t_{(11)}>2.021) =0.0342$

If we compare the the p value with the significance level provided $\alpha=0.1$ , we see that $p_v < \alpha$ , so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation :

x=test value 2017 , y = test value 2019

x: 28.7 32.1 29.6 30.5 31.9 30.9 32.3 33.1 29.6 30.8 31.1 31.6

y: 31.1 32.4 31.3 33.5 31.7 32.0 31.8 29.9 31.0 32.8 32.7 33.8

Solution to the problem

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \leq 0$

Alternative hypothesis: $\mu_y -\mu_x >0$

Because if we have an improvement we expect that the values for 2019 would be higher compared with the values for 2017

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: 2.4, 0.3, 1.7,3,-0.2, 1.1, -0.5, -3.2, 1.4, 2, 1.6, 2.2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{11.8}{12}=0.983$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.685$

We assume that the true difference follows a normal distribution. The 4th step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021$

The next step is calculate the degrees of freedom given by:

$df=n-1=12-1=11$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(11)}>2.021) =0.0342$

4 0

3 years ago

Help meeeeeeeeeeeeeee pls

WITCHER [35]

Answer:

Area = 15.36 ft

Step-by-step explanation:

First find the area of the dotted line triangle using the formula: base x height x 1/2, which would be 1.2ft x 12.8 x 1/2 = 7.68ft.

Next, add another triangle at the top, mirroring the shown triangle, of the parallelogram.

Then, find the area of the shape as though it were a rectangle. (base x height) - 2.4 x 12.8 = 30.72.

Next, multiply the area of the invisible triangle by 2, 7.68 x 2 = 15.36ft

and Finally, subtract the area of the rectangle shape of the the sum of the two invisible triangles, 30.72 - 15.36 = 15.36ft

The area of the parallelogram is 15.36 ft.

5 0

3 years ago

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

3 years ago

How do you do this??? 5x+9y+3x-2y this is a algebraic exercise plz help!!!

densk [106]

Combine like terms.

4. 5x+9y+3x-2y
8x+7y

7. 22u-6u+4t+4t-8u
8u+8t

10. 23m+4n-3m
20m+4n

6 0

3 years ago