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2 years ago

The table below shows the lengths and weights of six dolphins at an aquarium. Use the data from Pax and Snowflake to

Mathematics

2 answers:

Drupady [299]2 years ago

5 0

Answer:

The dolphin's weight that differs most from the value predicted by this model is JoJo's (31 lb of difference)

Step-by-step explanation:

Name Length (ft) Weight (lb)

Snowflake 8.2 549

Shadow 7.4 501

Corky 8.4 557

Pax 7.6 519

JoJo 5.8 460

Blaze 6.8 488

Use the data from Pax and Snowflake to create a linear model that predicts the weight (y) of a dolphin given its length (X). Which dolphin's weight differs most from the value predicted by this model?

Let's elaborate the model this way:

Snowflake's length - Pax's length = 8.2 - 7.6 = 0.6 ft

Snowflake's weight - Pax's weight = 549 - 519 = 30 lb

Ratio = 30/0.6 = 50 lb/ft

This means that for every foot of length, the dolphin has a variable weight of 50 lb. However, if we multiply the ratio by the actual length of the two dolphins, we will have these results:

Snowflake = 50 * 8.2 = 410

Pax = 50 * 7.6 = 380

What is the common difference in both cases?

Snowflake = Actual weight - Ratio * Length = 549 - 410 = 139

Pax = Actual weight - Ratio * Length = 519 - 380 = 139

The common difference is 139. Therefore, the linear model that predicts the weight (y) of a dolphin given its length (X) is:

y = 139 + 50x

Now, let's use the linear equation or model to calculate the weight of the remaining four dolphins, as follows:

Blaze y = 139 + 50 * 6.8 = 479 ⇒ difference of 9 lb (488 - 479)

JoJo y = 139 + 50 * 5.8 = 429 ⇒ difference of 31 lb (460 - 429)

Corky y = 139 + 50 * 8.4 = 559 ⇒ difference of 2 lb (559 - 557)

Shadow y = 139 + 50 * 7.4 = 509 ⇒ difference of 8 lb (509 - 501)

The dolphin's weight that differs most from the value predicted by this model is JoJo's (31 lb of difference)

Viefleur [7K]2 years ago

3 0

Answer:

jojo on edge

Step-by-step explanation:

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