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2 years ago

Use a composite figure to estimate the area of the figure. The grid has squares with side lengths of 1.5 cm.

Mathematics

1 answer:

ipn [44]2 years ago

4 0

The area of the composite figure in which the length of squares is 1.5 cm is 49.1 $cm^{2}$

Given that the grid has squares with side lengths of 1.5 cm.

We are required to find the area of the composite figure.

The area of the composite figure is approximate to the area of the rectangle plus the are of the semicircle at the top of the rectngle plus the semicircle at the right side of the rectangle.

Finding the area of the rectangle.

A=(3*1.5)(4*1.5)=27 $cm^{2}$

Finding the area of the semicircle at the top of the rectangle.

A=1/2 π $(2*1.5)^{2}$ =4.5 π $cm^{2}$

Finding the area of the semicircle at the right of the rectangle.

A=1/2 π $(1.5*1.5)^{2}$

=2.53π $cm^{2}$

Total area =27+4.5π+2.53π

=27+7.03π

Use π=3.14

=27+7.03*3.14

=49.1 $cm^{2}$

Hence the area of the composite figure in which the length of squares is 1.5 cm is 49.1 $cm^{2}$ .

Learn more about area at brainly.com/question/25965491

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The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv

Rina8888 [55]

Answer:

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$

So on this case the 95% confidence interval would be given by (12.150;16.690)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=14.42$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=6.95$ represent the population standard deviation

n=36 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$

So on this case the 95% confidence interval would be given by (12.150;16.690)

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3 years ago

Describe how you would find the of 24 36 and 13

erik [133]

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Solve the compound inequality.<br><br> 9 – 4x ≥ 5 or 4(–1 + x) – 6 ≥ 2

Y_Kistochka [10]

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3 years ago

The cars access code has four digits first digit cannot be zero and the last digit must be even

arlik [135]

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3 years ago

PLZ HELP, AND PLZ EXPLAIN

shutvik [7]

Answer:

Step-by-step explanation:

To make it easy let's start by organizing our information :

AC=12 AND BD=8
ABCD is a rhombus
K and L are the midpoints of sides AD and CD
we notice that the rhombus ABCD is divided into four right triangles

What do you think of when you hear a right triangle ?

The pythagorian theorem !

AC and BD are khown so let's focus on them .

If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?

Simply because they are the diagonals of a rhombus .

ow let's apply the pythagorian theorem :

(AC/2)² + (BD/2)² = BC²
6²+4²=52
BC²= 52⇒ $\sqrt{52}$ =BC

Now we khow that : AB=BC=CD=AD= $\sqrt{52}$

This isn't enough . Let's try to figure out a way to calculate the length of KL wich is the base of the triangle

KL is parallel to AC
k is the midpoint of AD and L of DC

I smell something . yes! Thales theorem

KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE

KL/12= $\sqrt{52}$ /2* $\sqrt{52}$
KL/12=1/2⇒ KL=6

Now we have the length of the base kl

Now the big boss the height :

notice that you khow the length of KL
BD crosses kl from its midpoint and DL = $\sqrt{52}$ /2

What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D

DL²=(KL/2)²+D²
52/4= 9+ D²
D² = 52/4-9 +4 SO D=2

now the height of the trigle is H= BD-D= 8-2=6

NOw the area of the triangle is :

A=(KL*H)/2 ⇒ A= (6*6)/2=18

THE ANSWER IS 18 SQ.UN

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3 years ago