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Harman [31]
3 years ago
5

Q: A: If two sides of a triangle are 3 cm and 5 cm in length, which could NOT be the measure of the third side? A) 2 cm B) 3 cm

C) 4 cm D) 5 cm
Mathematics
1 answer:
Zina [86]3 years ago
3 0

Answer:

A) 2cm

Step-by-step explanation:

According to the Triangle Inequality Theorem, the sum of the measure of any two sides of a triangle MUST be larger than the measure of the third side.  In this case, with sides 3cm and 5cm, the third side would have to be at least 3cm in order for the smallest two sides to be larger than the third side:

3cm + 3cm = 6 cm > 5cm

However, if the third side was only 2cm, then the two smaller sides would be 5cm, which is only equal to and not larger than the third side:

3cm + 2cm = 5cm

Any other measurements larger than 3cm, when added to the given side of 3cm, would be larger than 5cm and would form a triangle.  So, the only measure that would NOT form a triangle would be 2cm.

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3 years ago
US average math SAT scores follow a normal distribution with a mean of 505 and a standard deviation of 112. A sample of 64 enter
Hitman42 [59]

Answer:

The claim that the scores of UT students are less than the US average is wrong

Step-by-step explanation:

Given : Sample size = 64

           Standard deviation = 112

           Mean = 505

           Average score = 477

To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.

Solution:

Sample size = 64

n > 30

So we will use z test

H_0:\mu \geq 477\\H_a:\mu < 477

Formula : z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

z=\frac{505-477}{\frac{112}{\sqrt{64}}}

z=2

Refer the z table for p value

p value = 0.9772

α=0.05

p value > α

So, we accept the null hypothesis

Hence The claim that the scores of UT students are less than the US average is wrong

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3 years ago
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~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$100000\\ P=\textit{original amount deposited}\\ r=rate\to 6.6\%\to \frac{6.6}{100}\dotfill &0.066\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semiannually, thus two} \end{array}\dotfill &2\\ t=years\dotfill &20 \end{cases}

100000=P\left(1+\frac{0.066}{2}\right)^{2\cdot 20}\implies 100000=P(1.033)^{40} \\\\\\ \cfrac{100000}{1.033^{40}}=P\implies 27288.97\approx P

7 0
2 years ago
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