All categories

2 years ago

2 1/6 divided by 3/4

Mathematics

2 answers:

Allushta [10]2 years ago

8 0

I’m pretty sure it’s .18 sorry if I’m wrong

Gekata [30.6K]2 years ago

4 0

Answer:

Exact form = 26/9

Decimal form = 2.8

Mixed number form = 2 8/9

Step-by-step explanation:

Hope this helps

You might be interested in

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

Gayle installed a rectangular section of hardwood flooring measuring 12 ft by 12 ft in her family room. She plans on increasing

icang [17]

The ansa is 256=(12+x)(12+x)

6 0

2 years ago

find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large

katrin2010 [14]

A generic odd number can be written as

$2k+1,\quad k \in \mathbb{Z}$

Since there is an odd number every two numbers, three consecutive odd numbers will be

$2k+1,\quad 2k+3,\quad 2k+5$

Now let's make up the equations: the sum of the first two is

$(2k+1)+(2k+3)$

And 27 less than 3 times the largest is

$3(2k+5)-27$

These two must be the same, so we have

$(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3$

Subtracting 4k and 3 from both sides gives

$1=2k \iff k=\dfrac{1}{2}$

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.

7 0

3 years ago

A random sample of 5120 permanent dwellings on an entire reservation showed that 1627 were traditional hogans.

krek1111 [17]

Answer:

a) $\hat p=\frac{1627}{5120}=0.3178$

b)The 99% confidence interval would be given by (0.301;0.355)

Step-by-step explanation:

1) Notation and definitions

$X=1627$ number of permanent dwellings on the entire reservation that are traditional hogans.

$n=5120$ random sample taken

$\hat p=\frac{1627}{5120}=0.318$ estimated proportion of permanent dwellings on the entire reservation that are traditional hogans.

$p$ true population proportion of permanent dwellings on the entire reservation that are traditional hogans.

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Part a

The point of estimate for p=the proportion of all permanent dwellings on the entire reservation that are traditional hogans is given by:

$\hat p=\frac{1627}{5120}=0.3178$

Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: