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3 years ago

Will an object 20ft below sea level have a different absolute value than an object 20ft above sea level .. explain

Mathematics

1 answer:

trapecia [35]3 years ago

6 0

No, the objects below the sea and above both have the absolute value of 20 ft.

The absolute value is how far the number is away from zero- it doesn't matter if it is in the positive or negative direction

For example, if you drove 20 miles from your house in one direction while your friends drove 20 miles in the other direction- both of the distances of you and your friend from the your house is 20 miles.

Distance is a great example of the absolute value because it is always positive. It doesn't matter what direction, it is always 20 miles from your house.

You might be interested in

Does this set of ordered pairs represent a function? Why or why not? {(–1, 5), (0,-3), (2,7), (4,0), (7, 5)}

Kisachek [45]

Yes, it does. Because these points do respect the "law" that for each "x" there is only one "y" that corresponds to it

8 0

3 years ago

Read 2 more answers

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago

A cashier has 26 bills consisting of three times as many ones as fives, one less ten than fives, and the rest twenties. If the v

monitta

Let number of fives = x then number of ones = 3x and number of tens = x - 1

so we can create the equation

x + 3x + x-1 + y = 26 where y = number of twenties
so
5x + y = 27

also we have the equation

5x + 3x + 10(x - 1) + 20y = 120

18x + 20y = 130..................................(1)
5x + y = 27 multiply by -20:-
-100x - 20y = -540..............................(2)
Adding equation (1) and (2)
-82x = -410
x = 5, that is 5 fives

Now plug x = 5 into equation 1:-
18(5) + 20y = 130
20y = 40
y = 2 , that is 2 twenties

So the answer is there are (3x) = 15 ones , 5 fives, 4 tens and 2 twenties

7 0

4 years ago

61/2×(8/9÷13/18)+(3/4) of 31/5

liberstina [14]

Answer:

$\frac{10969}{260}$