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2 years ago

10

If you flip a coin and pick a number tile from a bag of beans containing 4 different numbered beans at the same time, how many p

ossible outcomes are there? A. 2 B. 4 C. 8 D. 12

1 answer:

Harman [31]2 years ago

4 0

Answer:

a

Step-by-step explanation:

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52.) Five times a number is 45

Rom4ik [11]

52). 5x=45 Divide each side of the equation by 5.

53). -3x = 12 Divide each side by -3.

54). x/4 = 10    Multiply each side by 4 .

55). x/3 = -8    Multiply each side by 3 .

1). x -10 = 12 . Add 10 to each side.

3). x + 8 = 16    Subtract 8 from each side.

5). 5 + x = 6 Subtract 5 from each side.

7). x - 4 = 9    Add 4 to each side.

Thank you for the 5 points. The crust and warm water are delicious.

4 0

2 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

Write the equation of the line passing through the two points. Show that this line is perpendicular to the given line.

il63 [147K]

Write the equation of the line passing through the two points. Show that this line is perpendicular to the given line.

(4,1) and (6,-5) y=1/3x-7

(1,-6) and (3,-10) y=1/2x-5

Please help, we learned this today and I’m really confused. How do you do it?

5 0

3 years ago

Find the value of x. Hint: the chords are congruent/equal (I will give 20 points)

Advocard [28]

$\huge \purple{\tt{Answer:}}$

If the chords are equal , then the measurement of both the chords will also be the same . That means ,

$\sf {6x-7 =35} \\\\\ \implies \sf {6x = 35 +7} \\\\\ \implies \sf {6x=42} \\\\\ \implies \sf{x= {\dfrac{\bcancel {42}^7}{\bcancel{6}}}} \\\\\ \implies \sf{x=7}$

5 0

2 years ago

NEED HELP!!! WILL MARK BRAINLIEST!!!

Tanzania [10]

Answer:

RTA= (x-2)·(x-11/2)/(x-2)(x-4)= (you can simplify again if you want by eliminating both (x-2)

(x-11/2)/(x-4)

Step-by-step explanation:

Ok we need to simplify the expression so:

x^2+3x-10= Bhaskara formula=

-3(±√9-4·1·(-10))/2·1=

X1=(-3+7/2)--> X1=2(R)---> (X-R)--->X-2

X2=(-3-7)/2 --> X2=11/2(R)---> (X-R)--->X-11/2

x^2-6x+8= Bhaskara formula=

6(±√36-4·1·8)/2·1=

X1=(6-2)/2=2--> X1=2(R)---> (X-R)--->X-2

X2=(6+2)/2=2--> X2=4(R)---> (X-R)--->X-4

so, The simplify expression is

(x-2)·(x-11/2)/(x-2)(x-4)=

8 0

2 years ago