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KIM [24]
3 years ago
13

Plz helps me I don’t get these I’m hoping for a smart math helper

Mathematics
2 answers:
fenix001 [56]3 years ago
8 0
1) since the probability of garden is 3/20, the total number of cards has to be a multiple of ten. 10 and 20 do not work, but 30 does. number 1 is 30

2) since the odds are 2/5, they're had to be 2/5*30=12

3) since the odds are3/10, there are 3/10*30=9

4)
mafiozo [28]3 years ago
6 0

probability of building + probability of garden + probability of monument = 1

                   2/5            +             3/10                 +            x                             = 1

                  2(2)            +             1(3)                  +         10(x)                           =10(1)

                    4              +              3                    +           10x                           =10

                                                                                        10x                           = 3

                                                                                           x                           =3/10

Total cards (c) x  probability of monument (3/10) = # of monument cards (9)  

3/10c = 9

c = 9(10/3)

c = 30 <em>Total postcards in the box</em>

Total cards (30) x  probability of building (2/5) = # of building cards (b)

30 x 2/5 = b

12 = b  <em>number of building postcards</em>

Total cards (30) x  probability of garden (3/10) = # of garden cards (g)

30 x 3/10 = g

9 = g  <em>number of garden postcards</em>

Answers: 30 postcards, 12 building cards, 9 garden cards

4)

Total cards (c) x  probability of building (2/5) = # of building cards (12 + 8)

c (2/5) = 20

c = 20(5/2)

c = 50

****************************************************************************************

Total cards (50) x  probability of garden (3/10) = # of garden cards (g)

50 (3/10) = g

15 = g

There were 9 garden cards and now there are 15

*****************************************************************************************

Total cards (50) x  probability of monument (3/10) = # of monument (m)

50 (3/10) = m

15 = m

There were 9 monument cards and now there are 15 cards

Answer:  6 more garden and 6 more monument postcards

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When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n whic
Elanso [62]

Answer:

95% Confidence interval for the mean

142.8 \leq\mu\leq158.4

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean of a finite population.

The error is multiplied by the following finite population correction factor:

cf=\sqrt{\frac{N-n}{N-1} }

The standard deviation can be estimated as

\sigma=\frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1} } =\frac{24.4}{\sqrt{32} }* \sqrt{\frac{200-32}{200-1} }=3.963

The 95% confidence interval has a z value of 1.96, so it becomes:

M-z*\sigma_c\leq\mu\leq M+z*\sigma_c\\\\150.6-1.96*3.963\leq\mu\leq 150.6+1.96*3.963\\\\ 142.8 \leq\mu\leq 158.4

5 0
3 years ago
Kaylen earned $14.87 selling cookies and $7.76 selling
____ [38]

Answer: $12.63

Step-by-step explanation:

$14.87 + $7.76 = $22.63

2pizzas x $5 = $10

$22.63 - $10 = $12.63

3 0
2 years ago
A store sells 3 videotapes for $4.99. find the unit rate
iogann1982 [59]
1.66333333 because u divide 4.99 by 3 to get the value of one
7 0
3 years ago
PLEEZZ
jonny [76]

4 < 4(6y - 12) - 2y       |use distributive property

4 < (4)(6y) + (4)(-12) - 2y

4 < 24y - 48 - 2y       |add 48 to both sides

52 < 26y      |divide both sides by 26

2 < y → y > 2

---------------------------------------------------------

4x + 3 < 3x + 6      |subtract 3 from both sides

4x < 3x + 3       |subtract 3x from both sides

x < 3

------------------------------------------------------------

-5r + 6 ≤ -5(r + 2)        |use distributive property

-5r + 6 ≤ (-5)(r) + (-5)(2)

-5r + 6 ≤ -5r - 10         |add 5r to both sides

6 ≤ -10     FALSE

No solution

-------------------------------------------------------------

-2(6 + s) ≥ -15 - 2s        |use distributive property

(-2)(6) + (-2)(s) ≥ -15 - 2s

-12 - 2s ≥ -15 - 2s        |add 2s to both sides

-12 ≥ -15     TRUE

All real numbers

-------------------------------------------------------------

3s + 6 ≤ -5(s + 2)        |use distributive property

3s + 6 ≤ (-5)(s) + (-5)(2)

3s + 6 ≤ -5s - 10         |subtract 6 from both sides

3s ≤ -5s - 16        |add 5s to both sides

8s ≤ -16         |divide both sides by 8

s ≤ -2

--------------------------------------------------------------

4/3 s - 3 ?

4 0
3 years ago
X/(x-2) + (x-1)/(x+1) = -1 <br><br> How to solve?
kirill115 [55]
Multiply entire equation by (x-2)(x+1) to get rid of the denominators That would lead to X(x+1)+(x-1)(x-2)=-1(x-2)(x+1)
Finally, using distributive property and foil, you would get x^2+x+x^2-3x+2=(-x^2+x+2).2x^2-2x+2=-x^2+x+23x^2-3x=0
3x(x-1)x=0 and x=1

7 0
3 years ago
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