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3 years ago

What is 1/3 divided by 5/6? Please help me don’t get it

2 answers:

7 0

Okay so this is the easiest way imo to do it.

Flip one of your fractions and then multiply.

1/3 * 6/5

1*6= 6
3*5= 15

6/15= 2/5

hammer [34]3 years ago

3 0

Answer:

=0.4

Step-by-step explanation:

1/3= 0.33

5/6=0.83

Then: 0.33/0.83 = 0.4

You can also do 6/15 which results from (1*6)=6 and (3*5)=15

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The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co

Alona [7]

Answer:

95% Confidence interval for the variance:

$3.6511\leq \sigma^2\leq 34.5972$

95% Confidence interval for the standard deviation:

$1.9108\leq \sigma \leq 5.8819$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

$s^2=2.89^2=8.3521$

The confidence interval for the variance is:

$\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}$

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

$\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128$

Then, the confidence interval can be calculated as:

$\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972$

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

$\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819$

6 0

3 years ago

Can anyone solve this, my teacher is still teaching through online learning and geometry is too confusing

malfutka [58]

Answer:

its too small to see

Step-by-step explanation:

7 0

3 years ago

Combine the following expressions. <br> <img src="https://tex.z-dn.net/?f=%5Csqrt%7B3y%5E2%7D%20%2B%204%5Csqrt%7B12y%5E2%7D%20-%

mote1985 [20]

ANSWER

$12y \sqrt{3}$

EXPLANATION

The given expression is

$\sqrt{3y^2} + 4\sqrt{12y^2} - y\sqrt{75}$

We identity and remove the perfect squares to obtain

$y\sqrt{3} + 16y\sqrt{3} - 5y\sqrt{3}$

We now observe that, the three terms are all similar.

We combine the similar terms to get:

$y\sqrt{3} + 16y\sqrt{3} - 5y\sqrt{3} = 12y \sqrt{3}$

4 0

3 years ago

Find the midpoint between these two points (-7,-6) and (2,5) PLEASE ANSWER!

QveST [7]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-7}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{2-7}{2}~~,~~\cfrac{5-6}{2} \right)\implies \left( \cfrac{-5}{2}~~,~~\cfrac{-1}{2} \right)\implies (-2.5,-0.5)$

4 0

3 years ago

Josie took a long multiple-choice, end-of-year vocabulary test. The ratio of the number of problems Josie got incorrect to the n

Digiron [165]

I think the answer is: she got 36 right and you would have 2 rectangles for the wrong answers in the tape diagram and 9 for the right.

hope I helped...

4 0

3 years ago