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3 years ago

What is 11/12 + -2/3

Mathematics

2 answers:

eimsori [14]3 years ago

8 0

Answer:

- 19/12 or - 1 7/12

Step-by-step explanation:

11/12 + ( - 8/12) = - 19/12 = - 1 7/12

Change the denominator for 12, and leave 11/12 alone, but multiply 4 to the numerator of - 2/3.

tensa zangetsu [6.8K]3 years ago

7 0

Answer:

11/12 + ( - 8/12) = - 19/12 = - 1 7/12

make the denominators the same so you can subtract easier

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4 years ago

A sample of 15 commuters in Chicago showed the average of the commuting times was 33.2 minutes. If s = 8.3 minutes, find the 95%

OleMash [197]

Answer:

The 95% confidence interval of the true mean.

(29.4261 ,36.9739)

Step-by-step explanation:

Step :- (i)

Given sample size 'n' =15

sample of the mean x⁻ = 33.2

The standard deviation of the sample 'S' = 8.3

95% of confidence intervals

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )$

Step:-(ii)

The degrees of freedom γ=n-1 = 15-1=14

The tabulated value t = 1.761 at 0.05 level of significance.

now substitute all possible values, we get

$(33.2 - 1.761\frac{8.3}{\sqrt{15} } ,33.2+ 1.761\frac{8}{\sqrt{15} } )$

After calculation , we get

(33.2-3.7739 , 33.2+3.7739

(29.4261 ,36.9739)

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8 0

3 years ago

PLZ HELP!!! Use limits to evaluate the integral.

Marrrta [24]

Split up the interval [0, 2] into n equally spaced subintervals:

$\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]$

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

$r_i=\dfrac{2i}n$

where $1\le i\le n$ . Each interval has length $\Delta x_i=\frac{2-0}n=\frac2n$ .

At these sampling points, the function takes on values of

$f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}$

We approximate the integral with the Riemann sum:

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3$

Recall that

$\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4$

so that the sum reduces to

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}$

Take the limit as n approaches infinity, and the Riemann sum converges to the value of the integral:

$\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}$

Just to check:

$\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28$

4 0

3 years ago

What is 11/12 + -2/3​

What is 11/12 + -2/3