DESPERATE FOR HELP! BRAINLIEST FOR FIRST RIGHT ANSWER!

State true or false. (i) Cube of any odd number is even. F (ii) A perfect cube does not end with two zeros.T (iii) If square of

lilavasa [31]

i. Cube of any odd number is even. False

ii. A perfect cube does not end with two zeros. True

iii. If square of a number ends with 5, then its cube ends with 25. False

iv. There is no perfect cube which ends with 8. False

v. The cube of a two digit number may be a three digit number. False

vi. The cube of a two digit number may have seven or more digits. False

vii. The cube of a single digit number may be a single digit number. True

<h3>Reasons </h3>

i. Cube of any odd number is even. This statement is false because the cube of odd numbers are also odd

ii. A perfect cube does not end with two zeros. This statement is True because the cube of a two digit number cannot be a 3 digit number and therefore cannot end with two zeros.

iii. If square of a number ends with 5, then its cube ends with 25. This statement is False

Using the illustration

35^2 = 35 x 35= 1225 , its square ends with 5

35^3= 35 x 35 x 35= 42875 which ends with 75

iv. There is no perfect cube which ends with 8. This statement is False.

Using the illustration, the cube of 2 is given thus

2 * 2 * 2 = 8

The cube of 2 ends with 8

v. The cube of a two digit number may be a three digit number. This statement is False.

Using the illustration, 10 is a two digit number

10 * 10 * 10 = 1000

The cube of 10 which is a two digit number is not a three digit number and same applies to all two digit numbers

vi The cube of a two digit number may have seven or more digits. This statement is False because the cube of a two digit number can only be a four digit number.

vii. The cube of a single digit number may be a single digit number. This statement is True because the cube of a single digit number can be a single, two or three digit number.

Learn more about cube numbers here:

https://brainly.in/question/6133463

#SPJ1

6 0

2 years ago

If the confidence level is 90%, find the Margin of sampling error. The population is normally distributed, the sample size is 15

charle [14.2K]

Answer:

$ME = 1.761* \frac{5}{\sqrt{15}}=2.273$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X= 75$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=5 represent the sample standard deviation

n=15 represent the sample size

Solution to the problem

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=15-1=14$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,14)".And we see that $t_{\alpha/2}=1.761$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$

And if we replace the values we got:

$ME = 1.761* \frac{5}{\sqrt{15}}=2.273$

6 0

3 years ago

A polymer is manufactured in a batch chemical process. Viscosity measurements show that it is approximately normally distributed

PilotLPTM [1.2K]

Answer: (751.05, 766.95)

Step-by-step explanation:

We know that the confidence interval for population mean is given by :-

$\overline{x}\pm z*\dfrac{\sigma}{\sqrt{n}}$ ,

where $\sigma$ =population standard deviation.

$\overline{x}$ = sample mean

n= sample size

z* = Two-tailed critical z-value.

Given : $\sigma= 20$

n= 42

$\overline{x}=759$

We know that from z-table , the two-tailed critical value for 99% confidence interval : z* =2.576

Now, the 99% confidence interval around the true population mean viscosity :-

$759\pm (2.5760)\dfrac{20}{\sqrt{42}}\\\\=759\pm (2.5760)(3.086067)\\\\=759\pm7.9497=(759-7.9497,\ 759+7.9497)\]\\=(751.0503,\ 766.9497)\approx(751.05,\ 766.95)$

∴ A 99% confidence interval around the true population mean viscosity : (751.05, 766.95)

3 0

4 years ago

Factor the expression. 35x-28xy=

zalisa [80]

Answer:

7x(5-4y)

Step-by-step explanation:

35x-28xy

7x(5-4y)

4 0

2 years ago

Read 2 more answers

Can someone please help me with these questions? I’m desperate! As always, I will give brainliest . I promise.

professor190 [17]

Since the centre of dilatation is the origin, we only require to multiply each of the coordinate points by the scale factor 3

(- 2, 1 ) → ( - 6, 3 )

(- 2, 3 ) → (- 6, 9 )

(1, 2 ) → (3, 6 )

(1, 3 ) → (3, 9 )

5 0

3 years ago