What is one hundred equal parts

1 answer:

3 0

If you have 100 things are the same in every way, that would be 100 equal parts. So if you have 100 of the same nails, that's 100 equal parts.

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The wildlife department has been feeding a special food to rainbow trout fingerlings in a pond. based on a large number of obser

Liono4ka [1.6K]

The normal distribution curve for the problem is shown below

We need to standardise the value X=405.5 by using the formula

$z-score= \frac{X-Mean}{Standard Deviation}$
$z-score= \frac{405.5-402.7}{8.8} =0.32$

We now need to find the probability of z=0.32 by reading the z-table

Note that z-table would give the reading to the left of z-score, so if your aim is to work out the area to the right of a z-score, then you'd need to do:

$P(Z\ \textgreater \ z)=1 - P(Z\ \textless \ z)$

from the z-table, the reading $P(Z\ \textless \ 0.32)$ gives 0.6255

hence,
$P(Z\ \textgreater \ 0.32)=1-0.6255=0.3475$

The probability that the mean weight for a sample of 40 trout exceeds 405.5 gram is 0.3475 = 34.75%

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2 years ago

Examine the rotation. Which best describes point D?

Nadusha1986 [10]

We can’t see the rotation

6 0

2 years ago

Read 2 more answers

Helppp Urgebttt!!!

Alenkasestr [34]

C because there are 60 seconds in a minute so if you multiply 600 x 60 you get 36000

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2 years ago

Use the figure to decide the type of angle pair that describes 6 and 4

kumpel [21]

It’s corresponding angles

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2 years ago

For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

7 0

3 years ago