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Contact [7]
3 years ago
10

What is the 15th term of the sequence A(n) = 50 - 3(n-1) ?

Mathematics
1 answer:
Alexus [3.1K]3 years ago
7 0

Answer:

A(15)=8

Step-by-step explanation:

Given that,

A(n)=50 - 3(n-1)

We need to find the 15th term of the sequence. Put n = 15 in the above sequence so that,

A(15)=50 - 3(15-1)\\\\A(15)=50-3(14)\\\\A(15)=50-42\\\\A(15)=8

So, the 15th term of the sequence is 8.

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5 0
3 years ago
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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