All categories

3 years ago

Need help for 3 2/3 - 1/2

Mathematics

2 answers:

sp2606 [1]3 years ago

7 0

The answer for 3 2/3 - 1/2 is 3.16 or 3 1/6

const2013 [10]3 years ago

3 0

3 2/3 - 1/2
= 11/3 - 1/2
= 22/6 - 3/6
= 19/6
= 3.1666666667

You might be interested in

Whoever give me the answer to this i will mark them as the Brainest if x = 5,y = 2 and z = 3 then xz ( 4y- 2z) is A. 3445 B. 210

max2010maxim [7]

x= 5 x z= 3 = 15 (4y - 2z) = 60 - -30 = 30 so its C

8 0

3 years ago

When sizes of pizzas are quoted in inches, the number quoted is the diameter of the pizza. A restaurant advertises an 8-inch “pe

mariarad [96]

Answer:

yes because if they want a bigger pizza or more they can just buy 2 eight inch pizza

Step-by-step explanation:

pizza is lit

8 0

3 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

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".

Solution to the problem

The population proportion have the following distribution

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

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

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.

$z_{\alpha/2}=2.58$

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0

3 years ago

Write in decimal from round to the nearest thousandth

Sveta_85 [38]

Divide each one, ex: 2÷12=2/12

6 0

3 years ago

The dimensions of a rectangular garden were 5m by 12m. Each dimensions was increased by the same amount. the garden has an area

Alex_Xolod [135]

Let's say, the amount that each side was increased was "a"
so, it was 5 wide and 12 long, now it's 5+a wide and 12+a long

we know the area is 120, so whatever "a" is, we know that

(5+a)(12+a) = 120

so $\bf (5+a)(12+a) = 120\implies 60+17a+a^2=120
\\\\\\
a^2+\underline{17} a\underline{-60}=0\qquad \qquad 
\begin{cases}
20\cdot -3\implies \underline{-60}\\\\
20-3\implies \underline{17}
\end{cases}$

use those factors, solve for "a"

4 0

3 years ago