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2 years ago

How many 5/6 in 10? Use models

1 answer:

4 0

10÷5/6
When dividing a number by a fraction, it is the same as multiplying by its reciprocal (the fraction flipped over)
so..
10÷5/6= 10×6/5= 60÷5= 12

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The value pie/2 is a solution for the equation 2sin^2x-sinx-1=0<br><br> true or false

Rasek [7]

Answer:

Yes, $\frac{\pi }{2}$ is a solution

Step-by-step explanation:

Recall that $sin(\frac{\pi }{2})=1$ . By replacing $\frac{\pi }{2}$ in the given trigonometric equation, we get 0 (zero) as a result, so this value is a solution:

$2*(sin(\frac{\pi }{2} ))^2-sin(\frac{\pi }{2})-1= 2*(1)^2-1-1=2-2=0$

6 0

3 years ago

Please Help???!!! I'm a 7th grader and need help! will give 20 points!

Ganezh [65]

Answer:

3. d= 20mm

4. r= 8 in

5. d= 14 mi

6. r= 11 yd

Step-by-step explanation:

Radius is half of diameter.

Example:

What is the diameter of a circle with a radius of 4?

If r = 4, d = 8

Do radius times 2.

Example 2:

What is the radius of a circle with a diameter of 10?

D = 10, so r = 5

Do diameter divided by 2.

5 0

3 years ago

Waht is 8/5 times 7/4 pleeaaaaseeee help LAST QUESTION PLZ

saveliy_v [14]

Answer:

below

Step-by-step explanation:

improper fraction: 14/5

as a decimal: 2.8

proper fraction: 2 4/5

8 0

2 years ago

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.