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Alex73 [517]
3 years ago
7

!!20 points!! I need help! I forgot how to solve a two step problem and I need a answer -6h -6=30​

Mathematics
2 answers:
Rudiy273 years ago
6 0

Answer:

h = -6

Step-by-step explanation:

BlackZzzverrR [31]3 years ago
3 0

Answer:

h = -6

Step-by-step explanation:

-6h -6=30

      +6 +6

-6h = 36

-6h/-6 = 36/-6

h = -6

Brainliest plz

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Which equation represents a circle with a center at (-4, 9) and a diameter of 10 units?
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(x + 4)2 + (y - 9)2 = 25

Step-by-step explanation:

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3 years ago
if log 2 = a, log 3 = b, and log 7 = c what is log 490 in terms of a, b and c and you can have integer coefficients
dsp73

Answer:

Answer is 2×c + log [10^{a} + 10^{b}] +a

Step-by-step explanation:

      log 2 = a ,log 3 = b ,  log 7 = c

      Using formula

      <em>log d×e = log d + log e</em>

<em>       log d^{a} = a×log d</em>

<em>       log 5 =  log [10^{a} + 10^{b}]</em>

      log 490 in terms of a , b and c and you can have integer coefficients

      log 490

       = log 7^{2}×5×2

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5 0
3 years ago
What is the probability that more than twelve loads occur during a 4-year period? (Round your answer to three decimal places.)
Nataly_w [17]

Answer:

Given that an article suggests

that a Poisson process can be used to represent the occurrence of

structural loads over time. Suppose the mean time between occurrences of

loads is 0.4 year. a). How many loads can be expected to occur during a 4-year period? b). What is the probability that more than 11 loads occur during a

4-year period? c). How long must a time period be so that the probability of no loads

occurring during that period is at most 0.3?Part A:The number of loads that can be expected to occur during a 4-year period is given by:Part B:The expected value of the number of loads to occur during the 4-year period is 10 loads.This means that the mean is 10.The probability of a poisson distribution is given by where: k = 0, 1, 2, . . ., 11 and λ = 10.The probability that more than 11 loads occur during a

4-year period is given by:1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]= 1 - 0.571665 = 0.428335 Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283Part C:The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:Therefore, the time period that must be so that the probability of no loads

occurring during that period is at most 0.3 is given by: 3.3 years

Step-by-step explanation:

8 0
4 years ago
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