What is the value of the expression (Three-fourths) cubed?

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

Drupady [299]

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)

5 0

2 years ago

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Which of the following are true statements about any regular polygon? Check all that apply. A. It is a quadrilateral. B. All of

sesenic [268]

A regular polygon is a closed figure
All its angles are equal in measure.
and all its sides are congruent

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2 years ago

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A random sample of 9 wheels of cheese yielded the following weights in pounds has a sample mean of 20.90 and a sample variance o

Ronch [10]

Answer:

$2.002 \leq \sigma^2 \leq 11.365$

Step-by-step explanation:

1) Data given and notation

s represent the sample standard deviation

$s^2$ represent the sample variance

n=9 the sample size

Confidence=90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^9 (x_i -\bar x)^2}{n-1}}
The sample variance given was [tex]s^2=3.45$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=9-1=8$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,8)" "=CHISQ.INV(0.95,8)". so for this case the critical values are:

$\chi^2_{\alpha/2}=15.507$

$\chi^2_{1- \alpha/2}=2.732$

And replacing into the formula for the interval we got:

$\frac{(9)(3.45)}{15.507} \leq \sigma \frac{(9)(3.45)}{2.732}$

$2.002 \leq \sigma^2 \leq 11.365$

7 0

2 years ago

Nancy used 14 yards of fabric to make flags for one third of the drill team. How much fabric would Nancy need to make flags for

Kipish [7]

If all the flags use material at the same rate, Nancy will need 3 times the fabric to make 3 times the number of flags. Nancy needs 3×(14 yd) = 42 yd for the whole team.

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2 years ago

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3.52 A coin is tossed twice. Let Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses.

Rufina [12.5K]

Answer:

a) The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

b) P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

c) P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

Step-by-step explanation:

Number of head on first toss = Z

Total Number of heads on 2 tosses = W

% of head occurring = 40%

% of tail occurring = 60%

P ( head ) = 2/5 , P( tail ) = 3/5

a) Determine the joint probability distribution of W and Z

P( W =0 |Z = 0 ) = 0.6 P( W = 0 | Z = 1 ) = 0

P( W = 1 | Z = 0 ) = 0.4 P( W = 1 | Z = 1 ) = 0.6

P( W = 1 | Z = 0 ) = 0 P( W = 2 | Z = 1 ) = 0.4

The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

B) Marginal distribution of W

P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

C) Marginal distribution of Z ( pmf of Z )

P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

8 0

2 years ago