Which statement accurately describes the relationship between mass and inertia?

Look at the following numbers and their labels A,B,C, and D.

zhuklara [117]

Answer:

Third option listed

Step-by-step explanation:

To find where to plot each letter, we have to round to the nearest ten--called the tens place.

321 [tens place underlined]

when rounding to a certain place value [like tens, hundreds, etc.], we round the digit by rounding the digit to the right of it.

If the digit to the right (in this case, the ones place ; 321)

is less than 5, we keep the digit in the tens place the same (and write a 0 in the ones place)

if the digit is greater than or equal to 5, we round the digit up (increase it by 1, and then write 0 in ones place)

so, here are the letters rounded:

A 269 = 270 [9 is greater than 5]

B 317 = 320 [7 is greater than 5]

C 298 = 300 [8 is greater than 5, and we carry the 1 from the tens place]

D 327 = 330 [7 is greater than 5]

Now, to find each value on the number line, we can look for that number [below each tick/mark]

So, the third option listed is correct

hope this helps!!

3 0

2 years ago

The perimeter of the rectangle is 56 cm. Find the value of x. 10 cm 3x cm

andriy [413]

56-10=46. 46 divided by 3=15.3 with the 3 repeating

4 0

3 years ago

Read 2 more answers

The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide

kondor19780726 [428]

Answer:

99% confidence interval for the population standard deviation = (74.97 , 635.20).

Step-by-step explanation:

We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation

$\sigma$ = population standard deviation

n = sample of university professors = 5

Also, $s^{2}$ = $\frac{\sum (X-\bar X)^{2} }{n-1}$ = 144.5

So, 99% confidence interval for population standard deviation, $\sigma$ is;

P(0.2070 < $\chi^{2} __5_-_1$ < 14.86) = 0.99 {As the table of $\chi^{2}$ at 4 degree of freedom

gives critical values of 0.2070 & 14.86}

P(0.2070 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 14.86) = 0.99

P( $\frac{ 0.2070}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 14.86}{(n-1)s^{2} }$ ) = 0.99

P( $\frac{ (n-1)s^{2}}{14.86 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{0.2070 }$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{14.86 }$ , $\frac{ (n-1)s^{2}}{0.2070 }$ )

= ( $\frac{ (5-1) \times 144.5^{2}}{14.86 }$ , $\frac{ (5-1) \times 144.5^{2}}{0.2070 }$ )

= (5620.525 , 403483.092)

99% confidence interval for $\sigma$ = ( $\sqrt{5620.525}$ , $\sqrt{403483.092}$ )

= (74.97 , 635.20)

Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).

8 0

4 years ago

Construct a 90% confidence interval for the true mean using the FPCF. (Round your answers to 4 decimal places.) The 90% confiden

mezya [45]

Answer:

The answer is below

Step-by-step explanation:

Twenty-five blood samples were selected by taking every seventh blood sample from racks holding 187 blood samples from the morning draw at a medical center. The white blood count (WBC) was measured using a Coulter Counter Model S. The mean WBC was 8.636 with a standard deviation of 3.9265. (a) Construct a 90% confidence interval for the true mean using the FPCF. (Round your answers to 4 decimal places.) The 90% confidence interval is from to

Answer:

Given:

Mean (μ) = 8.636, standard deviation (σ) = 3.9265, Confidence (C) = 90% = 0.9, sample size (n) = 25

α = 1 - C = 1 - 0.9 = 0.1

α/2 = 0.1/2 = 0.05

From the normal distribution table, The z score of α/2 (0.05) corresponds to the z score of 0.45 (0.5 - 0.05) which is 1.645

The margin of error (E) is given by:

$E=z_{\frac{\alpha}{2} }*\frac{\sigma}{\sqrt{n} }\\ \\E=1.645*\frac{3.9265}{\sqrt{25} }=1.2918$

The confidence interval = μ ± E = 8.636 ± 1.2918 = (7.3442, 9.9278)

The 90% confidence interval is from 7.3442 to 9.9278

8 0

4 years ago

A baker displays 144 cookies on 9 identical platters. How many platters does the baker need to display 64 cookies? Enter your an

NISA [10]

Answer:

4 platters

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers