All categories

3 years ago

Find the greatest possible error for each measurement. 10 1/8 oz

Mathematics

1 answer:

scoray [572]3 years ago

3 0

<u>Answer:</u>

0.0005 oz

<u>Step-by-step explanation:</u>

Usually, the greatest number that is allowed for approximation, assuming that the number itself is obtained by approximation, is the greatest possible error of it.

It is normally half the place value of the last digit in a number.

Like here we have $10\frac{1}{8}$ oz which is equal to $10.125$ oz. The last digit is 5 which is at the thousandth place (0.001) so the greatest possible error for this would be its half.

$\frac{0.001}{2}$ = 0.0005 oz

You might be interested in

Is y2=x-1 a function i will give brainliest

adelina 88 [10]

I don’t think it issss

6 0

2 years ago

The length of the hypotenuse of a right triangle is 4.5 inches. The length of one of the legs is 2.7 inches. What is the length

cupoosta [38]

Answer:

B. 3.6 inches

3 0

3 years ago

Please help mee pleaseee

nalin [4]

Answer:

The exact answer in terms of radicals is $x = 5*\sqrt[3]{25}$

The approximate answer is $x \approx 14.62009$ (accurate to 5 decimal places)

===============================================

Work Shown:

Let $y = \sqrt[5]{x^3}$

So the equation reduces to -7 = 8-3y

Let's solve for y

-7 = 8-3y

8-3y = -7

-3y = -7-8 ... subtract 8 from both sides

-3y = -15

y = -15/(-3) ... divide both sides by -3

y = 5

-----------

Since $y = \sqrt[5]{x^3}$ and y = 5, this means we can equate the two expressions and solve for x

$y = 5$

$\sqrt[5]{x^3} = 5$

$x^3 = 5^5$ Raise both sides to the 5th power

$x^3 = 3125$

$x = \sqrt[3]{3125}$ Apply cube root to both sides

$x = \sqrt[3]{125*25}$

$x = \sqrt[3]{125}*\sqrt[3]{25}$

$x = \sqrt[3]{5^3}*\sqrt[3]{25}$

$x = 5*\sqrt[3]{25}$

$x \approx 14.62009$

4 0

3 years ago

Read 2 more answers

It is hard to calculate the volume of a mountain but several estimates put the volume of Mount Everest at around 2,413 cubic kil

scoray [572]

Answer:

About 609,000 Cowboy stadiums could fit inside of Mount Everest

Step-by-step explanation:

we have

The estimate volume of Mount Everest is at around $2,413\ km^{3}$

The Dallas Cowboys Stadium has a volume of $140,000,000\ ft^{3}$

step 1

Convert ft³ to km³

we know that

1 km=3,280.84 ft

$140,000,000\ ft^{3}=140,000,000*(1/3,280.84)^{3}=0.003964\ km^{3}$

step 2

To find how many Cowboy stadiums could fit inside of Mount Everest, divide the volume of Mount Everest by the volume of the Dallas Cowboys Stadium

$2,413/0.003964=608,729$

Round to the nearest Thousands

$608,729=609,000$

The volume of Mount Everest is about 609,000 times greater than the volume of the Dallas Cowboys Stadium

8 0

2 years ago

Increments of seven

BaLLatris [955]

Answer:

Specific Learning Outcomes:

Solve problems that involve finding powers of a number

Description of mathematics:

In this problem students work with powers of numbers and, as a consequence, come to understand what is happening to the numbers.

Students also see how an apparently enormous and difficult calculation can be broken down into manageable parts. The students should come to realise that there are only a limited number of unit digits obtained when 7 is raised to a power. Further, these specific digits 'cycle round' as the power of 7 increases. This cycle is 7, 9, 3, 1, 7, 9, …

The same is true of the digit in the tens place.

6 0

3 years ago