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erica [24]
3 years ago
8

a fisherman catches 2 fish the amberjack weighs 3 times as much as the snapper,altogether the fish weigh 44 pounds how much does

the snapper weigh.
Mathematics
1 answer:
Anna [14]3 years ago
8 0

Because of the fact that the amberjack is essentially three snappers, it can be shown as 3s

since one snapper is one s, and the amberjack is 3s, we can combine like terms

s+3s = 4s

because altogether, the fish weigh 44 pounds, 4s would equal 44 (shown as 4s = 44), from there, simply divide by 4 on both sides to get the weight of a snapper;

s = 11

Small bonus (finding an amberjack)

Since an amberjack is equal to three snappers, and a snapper is 11 pounds, an amberjack would be 33 pounds.

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<h3>SOH CAH TOA identity</h3>

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In the manufacturing of a chemical adhesive, 9% of all batches have raw materials from two different lots. This occurs when hold
natali 33 [55]

Answer:

Step-by-step explanation:

Hello!

In a chemical adhesive manufacturing company you can encounter the following situations:

The adhesive is made in batches.

9% of all batches have raw materials from two different lots.

5% of all batches with material from a single lot require reprocessing.

40% of all batches with the material of two or more lots require additional processing.

You have two events:

A: A batch is formed with two different lots.

B: The lot requires additional processing.

And you can define their complements:

A': A batch is formed with one lot.

B': The lot does not require additional processing.

a. This probability is given in the text "9% of all batches have raw materials from two different lots. ", symbolically:

P(A)= 0.09

b. If the probability of a batch containing two lots is 9%, then it's complimentary event has a probability equal to:

P(A')= 1 - 0.09= 0.91

c. P(B/A) It is a conditional probability and you can interpret it as "the lot requires additional processing given that the batch is formed with two lots"

This information is also given:

P(B/A)= 0.40

d. P(B/A') Is the probability of the lot requiring additional processing given that the batch is formed with one lot.

P(B/A')= 0.05

e. Now you have to calculate the probability that "the batch is made with two lots" and "the lot requires additional processing"

A and B aren't independent events, which means that the occurrence of one of them modifies the probability of occurrence of the other one in two replications of the experiment.

P(A ∩ B)= P(A) * P(B/A)

P(A ∩ B)= 0.09 * 0.40= 0.036

f. You have to calculate the probability of "batch is formed with two different lots" and "the lot does not require additional processing", these two events aren't independent.

You cannot calculate the probability of this intersection the same way you did in item e. because we have no information about the probability of P(B'/A). But the P(A) is equal to the summation of P(A ∩ B) and P(A ∩ B'), so:

P(A ∩ B')= P(A) - P(A ∩ B)= 0.09 - 0.036= 0.054

g. To calculate the probability of the event "The lot requires additional processing." you have to add P(A ∩ B)+P(A' ∩ B)

First calculate P(A' ∩ B) = P(A')* P(B/A')= 0.91*0.05= 0.0455

P(B)= P(A ∩ B) + P(A' ∩ B)= 0.036 + 0.0455= 0.0815

I hope you have a SUPER day!

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Sladkaya [172]

Answer:

-2   hope this helps!   :)

Step-by-step explanation:

-10 over 5 is -2

so a = -2

7 0
3 years ago
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