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All rectangles are quadrilaterals true or false Hurry

Lorico [155]

Answer:

Yes.

Step-by-step explanation:

A quadrilateral is any polygon with four sides. A rectangle, by definition, is a polygon with four sides and four right angles. Thus, every rectangle is a quadrilateral.

7 0

3 years ago

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Which would feed more people 4/6 of a grape or 3/6 of a watermelon

Westkost [7]

Hello Almighty11,
Picture a grape and almost all of it, thats how much 4/6 of a grape would be.

3/6 of a watermelon is half of a whole watermelon, so it would definitely be the 3/6 watermelon.
~Naterator
Please Thank If This Helped

4 0

3 years ago

Each earbud is 4/7 of an inch long. If you put 8 earbuds end to end, how long would they be from beginning to end?

Lisa [10]

Answer:

4 4/7

Step-by-step explanation:

4/7x8/1

4x8/7x1

32/7

4 4/7

6 0

3 years ago

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A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.57 and the standard deviation is $20.28.

son4ous [18]

Answer:

The 98% confidence interval for the mean purchases of all customers is ($37.40, $61.74).

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.01 = 0.99$ , so $z = 2.325$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.325*\frac{20.28}{\sqrt{20}} = 12.17$

The lower end of the interval is the mean subtracted by M. So it is 49.57 - 12.17 = $37.40.

The upper end of the interval is the mean added to M. So it is 49.57 + 12.17 = $61.74.

The 98% confidence interval for the mean purchases of all customers is ($37.40, $61.74).

3 0

3 years ago

Dinosaur fossils are often dated by using an element other than carbon, like potassium-40, that has a longer half life (in this

Firlakuza [10]

The amount of substance left of a radioactive element of half life, $t_{\frac{1}{2}}$ after a time, t, is given by:

$N(t)=N_0\left( \frac{1}{2} \right)^ \frac{t}{t_{ \frac{1}{2} }}$

Given that potassium-40 has a half life of approximately 1.25 billion years.

The number of years it will take for 0.1% of potassium-40 to remain is obtained as follows:

$0.1=100\left( \frac{1}{2} \right)^ \frac{t}{1.25}} \\ \\ \Rightarrow\left( \frac{1}{2} \right)^ \frac{t}{1.25}}=0.001 \\ \\ \Rightarrow\frac{t}{1.25}\ln\left( \frac{1}{2} \right)=\ln(0.001) \\ \\ \Rightarrow \frac{t}{1.25}= \frac{\ln(0.001)}{\ln\left( \frac{1}{2} \right)} =9.966 \\ \\ t=9.966(1.25)=12.5$

Therefore, the maximum age of a fossil that we could date using 40k is 12.5 billion years.

3 0

3 years ago