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agasfer [191]
3 years ago
6

Four sevenths plus five sixths

Mathematics
2 answers:
aliya0001 [1]3 years ago
7 0

Answer:

1.40476190476

Step-by-step explanation:

Anna71 [15]3 years ago
5 0

Answer:

\frac{59}{42} or 1\frac{17}{42}

Step-by-step explanation:

\frac{4}{7} + \frac{5}{6}    (These fractions are not like and we can't add them so we must convert them to like fractions by multiplying them to get a common denominator, the LCM of 7 and 6 is also their product which is 42, because these numbers are co-prime)

\frac{4*6}{7*6} = \frac{24}{42}, \frac{5*7}{6*7} = \frac{35}{42}    (We can add these now that they are like)

\frac{24 + 35}{42} = \frac{59}{42} or 1\frac{17}{42}

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15x^3+4x^2+14x+12
Multiply 3x by every value in the second bracket. Then multiply 2 by every value in the second bracket. Add like terms. (Only answers with the same square are like terms.)
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Vika [28.1K]

Answer:

Step-by-step explanation:

a). Ratio of OP and RT = \frac{OP}{RT}

                                     = \frac{12}{3.6}

                                     = \frac{10}{3}

b). Since ΔOPQ ~ ΔRTS, their corresponding sides will be proportional.

    \frac{OQ}{RS}=\frac{OP}{RT}

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    OQ = 15 cm

c). m∠P = m∠T = 60°

d). m∠Q = m∠S

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    m∠R + m∠S + m∠T = 180°

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3 0
3 years ago
The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest
Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)

Calculation the value from standard normal z table, we have,  

P(x < 5) =0.7287= 72.87\%

b) P( port handles 3 or more million tons of cargo per week)

P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)

Calculating the value from the standard normal table we have,

1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%

c)P( port handles between 3 million and 4 million tons of cargo per week)

P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%

P(3 \leq x \leq 4) = 23.7\%

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85  

= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85  

=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15  

Calculation the value from standard normal z table, we have,  

P( z \leq -1.036) = 0.15

\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0
3 years ago
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