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3 years ago

Hi guys please help me to solve the following question

Mathematics

1 answer:

Troyanec [42]3 years ago

6 0

Answer:

Step-by-step explanation:

To find the area, we multiply the length times the width to find the number of squares in each box.

A) 1 square down times 7 squares across is 1 x 7, which is 7.

B) 2 squares down times 6 squares across is 2 x 6= 12.

C) 3 squares down times 5 squares across is 3 x 5 = 15.

D) 4 squares down times 4 squares across = 4 x 4 =16.

16 is the biggest number here, so that is the greatest area!

If you enjoyed my answer, please give it a rating and a Thanks! If you think this answer was pretty brainly, please give it a Brainliest!

Have a great day!

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Mashcka [7]

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3 years ago

Lauren made the plan shown for proving that quadrilateral ABCD with

ch4aika [34]

Answer:

Step-by-step explanation:

It is given that the quadrilateral ABCD has AB ≅ CD and BC ≅ DA is a parallelogram, then in order to prove opposite angles of the parallelogram are equal, we take ΔABC and ΔADC,

AC=AC(Common)

AB=CD(given)

BC=AD(given)

Thus, by SSS rule, ΔABC ≅ ΔADC

By CPCT, ∠B=∠C

Also, from ΔABD and ΔBCD, we have

AB=CD(given)

BC=AD(given)

BD=BD(common)

Thus, by SSS rule, ΔABD ≅ ΔBCD

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Since, opposite angles are equal,therefore ABCD is a parallelogram.

Draw in diagonals AC and BD. The given information and the shared side AC along with the Reflexive Property can be used to prove ΔABC ≅ ΔADC by the SSS Congruence Postulate. Using CPCTC, ∠B=∠C.The same can be done for ΔABD ≅ ΔBCD using the given information and the shared side BD. This will lead to ∠A=∠C. Therefore, ABCD is a parallelogram because opposite angles are congruent.

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3 years ago

The owner of a construction company would like to know if his current work teams can build room additions quicker than the time

frez [133]

Answer:

The claim that the current work teams can build room additions quicker than the time allotted for by the contract has strong statistical evidence.

Step-by-step explanation:

We have to test the hypothesis to prove the claim that the work team can build room additions quicker than the time allotted for by the contract.

The null hypothesis is that the real time used is equal to the contract time. The alternative hypothesis is that the real time is less thant the allotted for by the contract.

$H_0: \mu=0\\\\H_1: \mu$

The significance level, as a storng evidence is needed, is α=0.01.

The estimated standard deviation is:

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As the standard deviation is estimated, we use the t-statistic with (n-1)=15 degrees of freedom.

For a significance level of 0.01, right-tailed test, the critical value of t is t=2.603.

Then, we calculate the t-value for this sample:

$t=\frac{x-\mu}{s_M} =\frac{-0.32-0}{0.05}= -6.4$

As the t-statistic lies in the rejection region, the null hypothesis is rejected. The claim that the current work teams can build room additions quicker than the time allotted for by the contract has strong statistical evidence.

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