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3 years ago

Help with #20 please!!!!!

Mathematics

1 answer:

shepuryov [24]3 years ago

7 0

Remember that things like this, always need to add up to 180°.

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Consider the following sets of sample data: A: $29,400, $30,900, $21,000, $33,200, $21,300, $24,600, $29,500, $22,500, $35,200,

Lana71 [14]

Answer:

CV for A = 21.8%

CV for B = 15.5%

Step-by-step explanation:

The formula for coefficient of variation is:

CV = Standard Deviation / Mean

So,

For A:

Mean = Sum/No. of items

= 391300/14

=$27950

and

SD = $6085.31

CV for A = 6085.31/27950 * 100

=21.77%

Rounding off to one decimal

CV for A = 21.8%

For B:

Mean = Sum/No. of items

= 43.58/11

=3.96

and

SD = 0.615

CV for B = 0.615/3.96 * 100

=15.53%

=15.5% ..

6 0

3 years ago

What is an equivalent expression for 3(x-12)? * O 3x-12 3x+12 O 3x-36 3x+36

Doss [256]

The correct answer is
3x - 36

7 0

3 years ago

Read 2 more answers

Help! given 2(x-9)=-10 prove x=4 with statements and reasons

Sedbober [7]

Step-by-step explanation:

Proof:

2 ( x - 9) = - 10 when x = 4

2 ( 4 - 9) = -10

8 - 18 = -10

-10 = -10 ..........hence proven

Reasons 

- When you replace X with 4, you can clearly see that the equation is equal to -10.

sorry I couldn't answer accurately, but I hope this helps

8 0

3 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago

Use a polynomial Identity to expand this squared binomial (x + 6)2

krek1111 [17]

Answer: x2 + 12x

Step-by-step explanation:

7 0

3 years ago