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3 years ago

Find the probability of rolling a 2 or a 5 on a standard dice. *

Mathematics

1 answer:

romanna [79]3 years ago

3 0

The probability of a 2 or a 5 is 50%

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Expressions that are equivalent to <br> –<br> 8(q+6).

Alona [7]

Answer:

q = -6

Step-by-step explanation:

8(q + 6)

= (8×q) (8×6)

= 8q + 48

8q = -48

q = -48/8

q = -6

8 0

4 years ago

Question 2: You want to donate 30% of the $400 earned at your yard sale to

lilavasa [31]

Answer:

10%= 40$ 30%=120$

Step-by-step explanation:

ok so u earn 400$ at a yard sale. you divide it by 100 400/100=4. ok so then you multiply four by 10=40 and that is 10% multiply it by 30 u get 120 or 30%

sorry for the bad spelling

4 0

4 years ago

Wren bought a baseball card last year for $2.25. This year the price dropped to $.45. What was the percent of decrease in the pr

const2013 [10]

The price:
- last year: $2.25
- this year: $.45
The decrease: $2.25 - $.45 = $1.80
2.25 : 1.80 = 100 : x
2.25 x = 180
x = 180 : 2.25
x = 80
Answer:
The percent of decrease is A ) 80%

4 0

3 years ago

Leona is having a birthday party on Sunday. Jeanette wants to know how many people will be coming. Leona tells her that there wi

konstantin123 [22]

Answer:77 and 87

Step-by-step explanation:

7×6=42 + 5+7=35/\ 35+42=77(first outcome)

7×6=42 + 5×9=45/\42+45=87 (second outcome.

7 0

3 years ago

The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The d

jek_recluse [69]

Answer:

50%

Step-by-step explanation:

68-95-99.7 rule

68% of all values lie within the 1 standard deviation from mean $(\mu-\sigma,\mu+\sigma)$

95% of all values lie within the 1 standard deviation from mean $(\mu-1\sigma,\mu+1\sigma)$

99.7% of all values lie within the 1 standard deviation from mean $(\mu-3\sigma,\mu+3\sigma)$

The distribution of the number of daily requests is bell-shaped and has a mean of 55 and a standard deviation of 4.

$\mu = 55\\\sigma = 4$

68% of all values lie within the 1 standard deviation from mean $(\mu-\sigma,\mu+\sigma)$ = $(55-4,55+4)$ = $(51,59)$

95% of all values lie within the 2 standard deviation from mean $(\mu-1\sigma,\mu+1\sigma)$ = $(55-2(4),55+2(4))$ = $(47,63)$

99.7% of all values lie within the 3 standard deviation from mean $(\mu-3\sigma,\mu+3\sigma)$ = $(55-3(4),55+3(4))$ = $(43,67)$

Refer the attached figure

P(43<x<55)=2.5%+13.5%+34%=50%

Hence The approximate percentage of light bulb replacement requests numbering between 43 and 55 is 50%

4 0

3 years ago