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Gnom [1K]
3 years ago
12

The answer to the question in the picture

Mathematics
1 answer:
wolverine [178]3 years ago
7 0
Add them all up to get it
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Please help me for brainliest, a heart, and 5 star rating please!!!
Karolina [17]
C=3.14 x 26
C=81.64

C=2(3.14 x 13)
C=81.64
6 0
3 years ago
(03.01 MC)
Kitty [74]

Answer:

hey i cat under stand without the graph

Step-by-step explanation:

4 0
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What is 2 1/6 - 3 4/9?
vodomira [7]
To solve the question, first factorize both fraction,which makes:

2 1/6 - 3 4/9
= 2/6 - 12/9
= 1/3-4/3
= -3/3
= -1

Hope ot helps!
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You start at (5, -1). You move up 5 units. Where do you end? On a coordinate plane
aliya0001 [1]

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Step-by-step explanation:

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2 years ago
Read 2 more answers
When fishing off the shores of Florida, a spotted seatrout must be between 10 and 30 inches long before it can be kept; otherwis
Mademuasel [1]

Answer:

There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so \mu = 22, \sigma = 4.

What is the probability that a fisherman catches a spotted seatrout within the legal limits?

They must be between 10 and 30 inches.

So, this is the pvalue of the Z score of X = 30 subtracted by the pvalue of the Z score of X = 10

X = 30

Z = \frac{X - \mu}{\sigma}

Z = \frac{30 - 22}{4}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 10

Z = \frac{X - \mu}{\sigma}

Z = \frac{10 - 22}{4}

Z = -3

Z = -3 has a pvalue of 0.00135.

This means that there is a 0.9772-0.00135 = 0.97585 = 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

3 0
3 years ago
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