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coldgirl [10]
2 years ago
12

Perform the indicated operations and reduce the answer.

Mathematics
1 answer:
morpeh [17]2 years ago
6 0
11/36
OR
.3055555555................
hope this helped
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Solve for x.<br> 0.2x – 3.2<br> = 1.8<br> 1<br> 5<br> X = ?
Hitman42 [59]

Answer:

X = 25

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable

5 0
1 year ago
Answer it right and i'll try to give you brainliest.
mario62 [17]

Answer:

384 cm²

Step-by-step explanation:

Since it is a square pyramid, the total surface area of the pyramid is the area of its square base plus the area of its 4 triangular faces.

\boxed{area \: of \: triangle =  \frac{1}{2} \times base \times height }

Area of a triangular face

= ½ ×12 ×10

= 60 cm²

\boxed{area \: of \: square = side \times side}

Base area of pyramid

= 12 ×12

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2 years ago
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olga_2 [115]

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Step-by-step explanation:

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2 years ago
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.

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2 years ago
If you roll a dice 12 times what is the probability of rolling a 2
Andrei [34K]
The probability of getting a two is 2/12. 
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3 years ago
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