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Thepotemich [5.8K]
3 years ago
5

The figure shows a kite inside a rectangle. Which expression represents the area of the shaded region? 2x2 4x2 6x2 8x2

Mathematics
2 answers:
Savatey [412]3 years ago
4 0
The picture in the attached figure

we know that
area of the <span>the shaded region=area of rectangle -area of a kite

area of rectangle=(3x+x)*(x+x)----> 4x*2x---> 8x</span>²

area of a kite=(1/2)*[d1*d2]
where d1 and d2 are the diagonals
d1=4x
d2=2x
so
area of a kite=(1/2)*[4x*2x]----> 4x²

area of the the shaded region=8x²-4x²-----> 4x²

the answer is 
4x²

Ipatiy [6.2K]3 years ago
4 0
The expression that represents the area of the shaded region is 4x². Mark the other answer Brainliest, not mine!
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The following table gives results from two groups of students who took a nonproctored test. Use a 0.01 significance level to tes
Nana76 [90]

Answer:

Original claim is \mu_{1} =\mu_{2}

Opposite claim is \mu_{1} \neq \mu_{2}

Null and alternative hypotheses:

H_{0}:\mu_{1} = \mu_{2}

H_{1} : \mu_{1} \neq \mu_{2}

Significance level: 0.01

Test statistic:

We can use TI-84 calculator to find the test statistic and P-value. The steps are as follows:

Press STAT and the scroll right to TESTS

Scroll down to 2-SampTTest... and scroll to stats.

Enter below information.

\bar{x_{1}}=70.29

Sx1=22.09

n_{1} = 30

\bar{x_{2}}=74.26

Sx2=18.15

n_{2} = 32

\mu_{1} \neq \mu_{2}

Pooled: Yes

Calculate.

The output is in the attachment.

Therefore, the test statistic is:

t=-0.78

P-value: 0.4412

Reject or fail to reject: Fail to reject

Final Conclusion: Since the p-value is greater than the significance level, we, therefore, fail to reject the null hypothesis and conclude that the there is sufficient evidence to support the claim that the samples are from populations with the same mean.

6 0
3 years ago
After a dilation with a center of (0, 0), a point was mapped as (4, –6) → (12, y). A student determined y to be –2. Evaluate the
crimeas [40]
That would not be right because 4 times 3 equals 12 and -6 times 3 equals -18.
8 0
3 years ago
A rectangle is 20 meters long and 20 meters wide. If the length is increased by 10% and the width is decreased by 10%, what is t
Solnce55 [7]

Consider that the initial length and width of the rectangle are given as,

\begin{gathered} l=20 \\ b=20 \end{gathered}

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\begin{gathered} L=(1+\frac{10}{100})\cdot l \\ L=(\frac{110}{100})\cdot20 \\ L=22 \end{gathered}

After the width is decreased by 10%, the new width (B) of the rectangle is calculated as,

\begin{gathered} B=(1-\frac{10}{100})\cdot b \\ B=(\frac{90}{100})\cdot20 \\ B=18 \end{gathered}

Then the area (A) of the new rectangle is calculated as,

\begin{gathered} A=L\times B \\ A=22\times18 \\ A=396 \end{gathered}

Thus, the new area of the rectangle is 396 square meters.

7 0
1 year ago
if two dice are thrown together , what is the probability of getting an even number on one die and odd number on the other die ?
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Answer:

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the probability of the die rolling an even number  =  the probability of the die rolling an odd number  =1/2

Let us name these  2  dice as  A  and  B

One die can roll an even and the other can roll an odd, in the following cases

A  rolls even and  B  rolls odd. The probability of this happening  =1/2×1/2

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∴  The required probability

=1/2×1/2+1/2×1/2

=1/4+1/4

=1/2

4 0
2 years ago
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