All categories

3 years ago

3x9=10 threes- Three =

2 answers:

5 0

Yes you are correct because if you do 10x3 that gives you thirty and then do 30-3 which is 27 and so since 3x9=27 the way you explained it will also get you to 27 and so thank you for the question.

Aleonysh [2.5K]3 years ago

5 0

3=9times3=10 do I kinda get it ?

You might be interested in

HELP ASAP!!Solve for x. Enter the solutions from least to greatest.

telo118 [61]

X^2 - 5x - 14 = 0

what multiplies to negative 14 and adds to negative 5? -7 and positive 2.

(x - 7) (x + 2) = 0

Lesser x = -2
Greater x = 7

4 0

3 years ago

Find the volume.<br> 15 points

Anton [14]

Answer:

the volume is 196

Step-by-step explanation:

17x8=136

6x10=60

136+60=196

3 0

3 years ago

State the common difference for the given A.P.3/2,1/2,-1/2,-3/2,......

a_sh-v [17]

Answer:

-1

Step-by-step explanation:

The common difference is the constant amount of change between the numbers in an arithmetic sequence.

To find the common difference, you take the last term and subtract it by the previous term.

$\frac{-3}{2} - \frac{-1}{2} =-1$

8 0

3 years ago

One urn contains one blue ball (labeled b1) and three red balls (labeled r1, r2, and r3). a second urn contains two red balls (r

Flura [38]

Answer:

12 possibilities

Step-by-step explanation:

In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.

The same thing occurs in the second urn, as all balls have different labels.

The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).

For the first urn, we have a combination of 4 choose 2:

C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities

For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.

In total we have 6 + 6 = 12 possibilities.

3 0

3 years ago

Read 2 more answers

A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with σ=0.0

oksian1 [2.3K]

<h2>Answer with explanation:</h2>

The confidence interval for population mean is given by :-

$\overline{x}\pm z^* SE$ (1)

, where $\overline{x}$ = sample mean

z* = critical value.

SE = standard error

and $SE=\dfrac{\sigma}{\sqrt{n}}$ , $\sigma$ = population standard deviation.

n= sample size.

As per given , we have

$\overline{x}=74.021$

$\sigma=0.001$

n= 15

It is known that ring diameter is normally distributed.

$SE=\dfrac{0.001}{\sqrt{15}}=0.000258198889747\approx0.000258199$

By z-table ,

The critical value for 95% confidence = z*= 1.96

A 99% two-sided confidence interval on the true mean piston diameter :

$74.021\pm (2.576) (0.000258199)$ (using (1))

$74.021\pm 0.000665120624$

$74.021\pm 0.000665120624\\\\=(74.021- 0.000665120624,\ 74.021+ 0.000665120624)\\\\=(74.0203348794,\ 74.0216651206)\approx(74.020,\ 74.022)$ [Rounded to three decimal places]

∴ A 99% two-sided confidence interval on the true mean piston diameter = (74.020, 74.022)

By z-table ,

The critical value for 95% confidence = z*= 1.96

A 95% lower confidence bound on the true mean piston diameter:

$74.021- (1.96) (0.000258199)$ (using (1))

$74.021- 0.00050607004=74.02049393\approx74.020$ [Rounded to three decimal places]

∴ A 95% lower confidence bound on the true mean piston diameter= 74.020

7 0

3 years ago