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3 years ago

Help me please with This

Mathematics

2 answers:

MAXImum [283]3 years ago

5 0

Answer:

Step-by-step explanation:

the answer is zero because this problem uses the commutative property meaning the numbers can come in any order and equal the same thing

stepladder [879]3 years ago

4 0

Answer:

I believe it's 0,hope that helps

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Consider the two triangles.

GuDViN [60]

Answer:

(A)Show that the ratios StartFraction U V Over X Y EndFraction , StartFraction W U Over Z X EndFraction , and StartFraction W V Over Z Y EndFraction are equivalent.

$\dfrac{UW}{XZ}=\dfrac{WV}{ZY}=\dfrac{UV}{XY}$

Step-by-step explanation:

In Triangles WUV and XZY:

$\angle VUW$ and \angle YXZ$ are congruent. \\\angle U W V$ and \angle X Z Y$ are congruent.\\ \angle U V W$ and \angle Z Y X$ are congruent.$

Therefore:

$\triangle UWV \cong \triangle XZY$

To show that the triangles are similar by the SSS similarity theorem, we have:

$\dfrac{UW}{XZ}=\dfrac{WV}{ZY}=\dfrac{UV}{XY}$

As a check:

$\dfrac{UW}{XZ}=\dfrac{40}{32}=1.25\\\\\dfrac{WV}{ZY}=\dfrac{60}{48}=1.25\\\\\dfrac{UV}{XY}=\dfrac{50}{40}=1.25$

The correct option is A.

4 0

3 years ago

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the significance $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=30.144$

$\chi^2_{1- \alpha/2}=10.117$

And replacing into the formula for the interval we got:

$\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

Part c

Now we just take square root on both sides of the interval and we got:

$23.818 \leq \sigma \leq 41.112$

5 0

3 years ago

Find the vertex for each equation:

tamaranim1 [39]

1. (−3,−1)

2. (3,−5)

3. (−3,−4)

5 0

3 years ago

Find the equation of an exponential function in the form y = ab^x, given the points (0, 3) and (2, 108/25). Please simplify your

lilavasa [31]

We have the equation:

$y=a\cdot b^x$

We know two points and we will use them to calculate the parameters a and b.

The point (0,3) will let us know a, as b^0=1.

$\begin{gathered} y=a\cdot b^x \\ 3=a\cdot b^0=a \\ a=3 \end{gathered}$

Now, we use the point (2, 108/25) to calcualte b:

$\begin{gathered} y=3\cdot b^x \\ \frac{108}{25}=3\cdot b^2 \\ 3\cdot b^2=\frac{108}{25} \\ b^2=\frac{108}{25\cdot3}=\frac{108}{3}\cdot\frac{1}{25}=\frac{36}{25} \\ b=\sqrt[]{\frac{36}{25}} \\ b=\frac{\sqrt[]{36}}{\sqrt[]{25}} \\ b=\frac{6}{5} \end{gathered}$

Then, we can write the equation as:

$y=3\cdot(\frac{6}{5})^x$

5 0

1 year ago

64^1/3=4 in logarithmic form

Marrrta [24]

$b^{a} = c
\\ \\In\ logarithmic\ form\ it\ can\ be\ written\ as
\\ \\log_{b}c=a.
\\ \\ So.\ in\ our\ case
\\ \\64^{\frac{1}{3}}=4
\\ \\ log_{64}4=\frac{1}{3}$

7 0

3 years ago

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