You have a stack of pennies without counting the pennies, how can you know if there is a odd or even number of them

2 answers:

8 0

Suppose that you have a stack of pennies. If you want to know without counting whether there is an odd or even number of pennies, then you have to split them into two parts, for example, placing one coin left and one coin right.

If the number of pennies is even, then no coins will left, if the number is odd, then one coin will be extra.

ratelena [41]3 years ago

7 0

You can put the pennies in 2 rows, if they match then they are even, if not they are odd

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Identify the corresponding word problem given the equation: x = 10.20 × 9

Inga [223]

10.20 x 9 = 91.8

( It's William and nine/eight friends, since they're dividing it among ALL people we count William in. )

A: William and nine friends divided the bill evenly. If each person paid $10.20, what was the total bill?

10.20 x 10 = 102

B: At a restaurant, William and eight friends divided the bill evenly. If each person paid $10.20, what was the total bill?

10.20 x 9 = 91.8

C: At a restaurant, William and nine friends divided the bill evenly. If the total bill was $91.80, what did each person pay?

91.80 x 10 = 918

D: At a restaurant, William and eight friends divided the bill evenly. If the total bill was $91.80, what did each person pay?

91.80 x 9 = 826.2

So, your answer is B

3 0

2 years ago

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The water diet requires you to drink two cups of water every half hour from the time you get up until you go to bed, but otherwi

Elan Coil [88]

Answer:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

Step-by-step explanation:

For this case we have the following info given:

Weight before diet 180 125 240 150

Weight after diet 170 130 215 152

We define the random variable $D = before-after$ and we can calculate the inidividual values:

D: 10, -5, 25, -2

And we can calculate the mean with this formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And the deviation with:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

And after replace we got:

$\bar D= 7, s_d = 13.638$

And the confidence interval for this case would be given by:

$\bar D \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}$

The degrees of freedom are given by:

$df = n-1= 4-1=3$

For the 95% of confidence the value for the significance is $\alpha=0.05$ and the critical value would be $t_{\alpha/2}= 3.182$ . And replacing we got: