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LuckyWell [14K]
3 years ago
11

5/6 divided by 3????? Pls help

Mathematics
2 answers:
Oduvanchick [21]3 years ago
6 0
5/6 divided by 3/1

5/6 *1/3
5/18
pentagon [3]3 years ago
3 0

\bold{Hey\ there!}

  • \bold{\frac{5}{6}\div3}
  • \bold{Covert:3\ to \rightarrow\frac{3}{1}}
  • \bold{\frac{5}{6}\div\frac{3}{1}}
  • \bold{Cross \ multiply(numerator\times denomenator)}
  • \bold{5\times1=5(\leftarrow numerator)}
  • \bold{6\times3=18(\leftarrow denominator)}
  • \boxed{\boxed{\bold{Your\ answer:\frac{5}{18}}}}\checkmark

\bold{Good \ luck\ on \ your \ assignment \ and\ enjoy} \bold{your \ day!}

~\frak{LoveYourselfFirst:)}

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Answer:

1st answer

Step-by-step explanation:

From pythagorus theorem.

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=  \sqrt{  {2}^{2}  +  {3}^{2}  }=   \sqrt{ {13}}

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3 years ago
An observational study determined that there is a strong correlation between eating breakfast and weight loss. Can it be determi
mylen [45]

A common misconception in statistics is confusing correlation with causation. If two events are correlated, it merely means that they share the same behaviour over time, but it doesn't imply in any way that those event are related by a common cause, or even worse, that one implies the other.

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7 0
3 years ago
There were 700 people in the auditorium. 60% of them were adults and the rest were children. how many adults were in the auditor
ollegr [7]

.60 \times 700 = 420 \\ 700 - 420 = 280
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5 0
3 years ago
Does anyone under stand this one ??
Annette [7]

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5 0
3 years ago
Read 2 more answers
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
3 years ago
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