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3 years ago

Can you make a obtuse triangle into a rectangle

1 answer:

7 0

Well,

If you can do anything to an obtuse triangle, you could add a side and make all the angles right angles.

If all you can do is alter the angles, you will not be able to do that because you will still have a triangle with three sides.

If all you can do is alter the number of sides, you will still not be able to do that because you will have some type of quadrilateral, not a rectangle.

In essence, you must be able to both change the angles and the number of sides in order to make any triangle a rectangle.

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If you roll two six-faced dice together, you will get 36 possible outcomes. 4 pts 1. List all possible outcomes of the experimen

bezimeni [28]

Given:

Two dice are rolled together.

Total number of possible outcomes.

To find:

The list of total possible outcomes.

The probability of getting a sum of 11 in these outcomes.

The probability of getting a sum less than or equal to 4.

The probability of getting a sum of 13 or more.

Solution:

If two dice are rolled together, then the total number of possible outcomes is 36 and list of total possible outcomes is

S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Sum of 11 in these outcomes = {(5,6),(6,5),(6,6)} = 3

The probability of getting a sum of 11 in these outcomes is

$P(\text{sum=11})=\dfrac{3}{36}$

$P(\text{sum=11})=\dfrac{1}{12}$

Therefore, the probability of getting a sum of 11 in these outcomes is $\dfrac{1}{12}$ .

Sum less than or equal to 4 = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} = 6

The probability of getting a sum less than or equal to 4 is

$P({sum\leq 4})=\dfrac{6}{36}$

$P({sum\leq 4})=\dfrac{1}{6}$

Therefore, the probability of getting a sum less than or equal to 4 is $\dfrac{1}{6}$ .

Sum of 13 or more = empty set because maximum sum is 12.

The probability of getting a sum of 13 or more is

$P(sum\geq 13)=\dfrac{0}{36}$

$P({sum\geq 13})=0$

Therefore, the probability of getting a sum of 13 or more is 0.

8 0

3 years ago

Dan built deck that is 5 feet long and 5 feet wide He built another deck that is 5 feet long and 7 wide. He built a third deck t

Elodia [21]

As the width gets larger so does the total area

5 0

3 years ago

the ratio of sadia's age to her father's age is 3:6. The sum of their age is 96 .What is sadia's age?

Mrrafil [7]

Answer:

sadia is 32

Step-by-step explanation:

sadia : father : total

3 6 9

Divide 96 by 9

96/9 = 32/3

Multiply each by 32/3

sadia : father : total

3*32/3 6*32/3 9*32/3

32 64 96

4 0

3 years ago

A retailer allowed 12% discount and sold a T-shirt at a loss of Rs 16. If he had sold it at 10% discount he would have gained Rs

Murrr4er [49]

Answer: Marked Prixe = Rs. 1800

Step-by-step explanation:

Let the marked price be M

1) A retailer allowed 12% discount and sold a T-shirt at a loss of Rs 16.

SP = 0.88M

CP - SP = 16

CP = 16 + SP

CP = 16 + 0.88M

2) If he had sold it at 10% discount he would have gained Rs 20.

SP = 0.9M

SP - CP = 20

0.9M = 20 + CP

CP = 0.9M - 20

16 + 0.88M = 0.9M - 20

0.9M - 0.88M = 20+16

0.02M = 36

36 = 0.02M

M = 1800

6 0

3 years ago

Suppose a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%. A cell phon

leva [86]

Answer:

We conclude that the actual percentage of households is equal to 30%.

Step-by-step explanation:

We are given that a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%.

Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Let p = proportion of households that have three cell phones NOT known.

So, Null Hypothesis, $H_0$ : p = 30% {means that the actual percentage of households is equal to 30%}

Alternate Hypothesis, $H_A$ : p $\neq$ 30% {means that the actual percentage of households different from 30%}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of households having three cell phones = $\frac{43}{150}$ = 0.29

n = sample of households = 150

So, the test statistics = $\frac{0.29-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }$

= -0.27

The value of z test statistic is -0.27.

Also, P-value of the test statistics is given by;

P-value = P(Z < -0.27) = 1 - P(Z $\leq$ 0.27)

= 1 - 0.6064 = 0.3936

Now, at 1% significance level the z table gives critical value of -2.58 and 2.58 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual percentage of households is equal to 30%.

4 0

3 years ago