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Lisa [10]

4 years ago

4" alt=" - \frac{9}{5} \div 4" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

raketka [301]4 years ago

3 0

Your answer would be negative 9/20

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Is number 1, 2, 3 and 4 correct?

ArbitrLikvidat [17]

\left[x _{2}\right] = \left[ 3\right][x2]=[3] totally answer

6 0

4 years ago

(this is a rhombus btw) for what value of x is the figure the given special parallelogram? HELP PLS, show work.

snow_lady [41]

Answer:

x = 2 or 3

Step-by-step explanation:

Diagonal of a rhombus bisects the vertex angles. So,

$x^{2} =5x-6$

$=> x^{2} -5x + 6 = 0$

Lets factorise the eqn.

$=> x^{2} - 2x - 3x + 6 = 0$

$=> x (x - 2) -3(x-2)$

$=> (x-2)(x-3) = 0$

Here x will have 2 values.

1) $x-2=0$

$=> x =2$

2) $x - 3 = 0$

$=> x = 3$

7 0

4 years ago

The mean cost of a meal for two in a mid-range restaurant in Tokyo is $40 (Numbeo.com website, December 14, 2014). How do prices

NNADVOKAT [17]

Answer:

a) $Me=2.02 \frac{6.827}{\sqrt{42}}=2.13$

b) So on this case the 95% confidence interval would be given by (30.53;34.79)

c) On this case the confidence interval not contains the price $40, so we can conclude that the prices for Hong Kong mid-range restaurants are significant less than the prices for mid range restaurants in Tokyo at 5 % of significance.

Step-by-step explanation:

1) 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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

2) Part a

The margin of error is given by:

$Me=t_{\alpha/2}\frac{s}{\sqrt{n}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=42-1=41$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,41)".And we see that $t_{\alpha/2}=2.02$

$Me=2.02 \frac{6.827}{\sqrt{42}}=2.13$

3) Part b

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=32.66$

The sample deviation calculated $s=6.827$

Now we have everything in order to replace into formula (1):

$32.66-2.02\frac{6.827}{\sqrt{42}}=30.532$

$32.66+2.02\frac{6.827}{\sqrt{42}}=34.788$

So on this case the 95% confidence interval would be given by (30.532;34.788)

4) Part d

On this case the confidence interval not contains the price $40, so we can conclude that the prices for Hong Kong mid-range restaurants are significant less than the prices in mid range restaurants in Tokyo at 5 % of significance.

8 0

4 years ago

If x is positive, is

vredina [299]

They are equal.

$\sqrt{\dfrac1{x^2}} = \dfrac1{|x|} = \dfrac1x$