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3 years ago

14

If you have thirty -five (35) members in an organization that meets nine (9) times a year, how often will a member have to host

a meeting?

1 answer:

Olegator [25]3 years ago

3 0

Answer:

<em>the</em><em> </em><em>host</em><em> </em><em>will</em><em> </em><em>continue</em><em> </em><em>the</em><em> </em><em>way</em><em> </em><em>they</em><em> </em><em>are</em><em> </em><em>prepared</em><em>. </em>

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Answer the statistical measures and create a box and whiskers plot for the following set of data. You may optionally click and d

ratelena [41]

sorry

look im trying to finish a challange u need premium if your going toask that big of a quietion my dude/dudet

3 0

3 years ago

Help please I am a bit confused on this.

Anit [1.1K]

Because there are 4 inside angles the sum of the four angles must equal 360 degrees.

Add the angles to equal 360:

4x + 3x + 2x + 3x = 360

Simplify:

12x = 360

Solve for x by dividing both sides by 12:

x = 360 /12

x = 30

Now you have x, solve for each angle:

ABC = 4x = 4 x 30 = 120 degrees.

BCD = 3x = 3 x 30 = 90 degrees.

CDA = 2x = 2x 30 = 60 degrees.

DAB = 3x = 3 x 30 = 90 degrees.

C. It's important to know that a four sides figure needs the inside angles when added together need to equal 360 degrees.

6 0

3 years ago

Read 2 more answers

Someone help me please

garri49 [273]

Answer:

42 seats

Step-by-step explanation:

This represents an arithmetic progression with the first term of a = 15 and common difference of d = 3.

<u>The tenth row is the 10th term:</u>

aₙ = a + (n - 1)d

<u>The row 10 has:</u>

a₁₀ = 15 + 9*3 = 15 + 27 = 42 seats

7 0

2 years ago

Read 2 more answers

(Small sample confidence intervals for a population mean) suppose you are taking a sampling of 15 measurements. you find that x=

Luda [366]

Answer:

The 99% confidence interval is $71.67 < \mu < 78.33$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 15$

The sample mean is $\= x = 75$

The standard deviation is $s = 5$

Given that confidence is 99% then the level of significance is mathematically represented as

$\alpha = 100 - 99$

$\alpha = 1\%$

$\alpha = 0.01$

Next we obtain the critical values of $\frac{ \alpha }{2}$ from the normal distribution table

The value is

$Z_{\frac{ \alpha }{2} } = 2.58$

Generally the margin for error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{ s}{ \sqrt{n} }$

=> $E = 2.58 * \frac{ 5}{ \sqrt{15} }$

=> $E = 3.3307$

The 99% confidence interval is mathematically represented as

$\= x -E < \mu < \= x +E$

=> $75 - 3.3307 < \mu$

=> $71.67 < \mu < 78.33$

3 0

3 years ago

of the last 15 contestants on a game show, 3 won a prize. Considering this data, how many of the next 20 contestants would you e

nadya68 [22]

You do 3/15 which is 0.2

then 0.2*20 which is 4

you can expect 4 to win a prize

hope this helps and you rate me good;)

4 0

3 years ago