A baby fish was 1.5 centimeters in length when it first hatched . The fish grew at a rate of 0.25 centimeter per week.
Which graph best shows this relationship between the length of the fish in centimeters , y, and the time in weeks, x?

A Baby Fish Was 1.5 Centimeters In Length When It First Hatched . The Fish Grew At A Rate Of 0.25 Centimeter

Answers

Answer 1

If you think about I would suggest B only because it makes more sense but I might be wrong..

Hope this helped a little!

Answer 2

The solution is Option A.

The equation of line is y = 0.25x + 1.5 , where the slope m = 0.25

What is an Equation of a line?

The equation of a line is expressed as y = mx + b where m is the slope and b is the y-intercept

And y - y₁ = m ( x - x₁ )

y = y-coordinate of second point

y₁ = y-coordinate of point one

m = slope

x = x-coordinate of second point

x₁ = x-coordinate of point one

The slope m = ( y₂ - y₁ ) / ( x₂ - x₁ )

Given data ,

Let the equation of line be represented as A

Now , the value of A is

The initial length of the baby fish is b = 1.5 cm

The baby fish grew at a rate of 0.25 centimeter per week

So , the slope of the line m = 0.25 cm

And , equation of a line is expressed as y = mx + b where m is the slope and b is the y-intercept

Substituting the values in the equation , we get

y = 0.25x + 1.5   be equation (1)

Hence , the equation of line is y = 0.25x + 1.5

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A Baby Fish Was 1.5 Centimeters In Length When It First Hatched . The Fish Grew At A Rate Of 0.25 Centimeter

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Past surveys reveal the 30% of tourists going to Las Vegas to gamble spend more than $1,000. The Visitor's Bureau of Las Vegas wants to update this percentage. The new study is to use the 90% confidence level. The estimate is to be within 1% of the population proportion. What is the necessary sample size?

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Answer:

The necessary sample size is of 5683.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

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Past surveys reveal the 30% of tourists going to Las Vegas to gamble spend more than $1,000.

This means that [tex]\pi = 0.3[/tex]

The estimate is to be within 1% of the population proportion. What is the necessary sample size?

A sample size of n is necessary. n is found when [tex]M = 0.01[/tex]. So

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.01 = 1.645\sqrt{\frac{0.3*0.7}{n}}[/tex]

[tex]0.01\sqrt{n} = 1.645\sqrt{0.3*0.7}[/tex]

[tex]\sqrt{n} = \frac{1.645\sqrt{0.3*0.7}}{0.01}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.645\sqrt{0.3*0.7}}{0.01})^2[/tex]

[tex]n = 5682.6[/tex]

Rounding up:

The necessary sample size is of 5683.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

The recursive formula tells you this is an arithmetic sequence with first term 51 and common difference 2. The explicit formula for such a sequence is ...

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Answer:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Answers

Answer:

Step-by-step explanation:

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Answers

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Answers

Answer:

t(s) is in the acceptance region. We accept H₀ we don´t have enough evidence to justify differences in the procedures

Step-by-step explanation:

Traditional procedure bring as μ  =  8,6 h

Sample Information:

Sample size   n  = 26

Sample mean  x  =  8,4

Sample standard deviation   s  = 1

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t(c) = - 1,708

Test Hypothesis

Null hypothesis                  H₀                 x =  μ

Alternative Hypothesis     Hₐ                 x ≠  μ

Alternative hypothesis indicates that the test should considered two tails

To calculate  t(s)

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t(s)  = ( 8,4  -  8,6 )/ 1/√26

t(s)  = - 0,2 *  5,09 / 1

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Comparing |t(s)| and |t(c)|

1,019 < 1,708     |t(s)| <  |t(c)|

t(s) is in the acceptance region. We accept H₀ we don´t have enough evidence to justify differences in the procedures

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Answer:

A(-5, -2), B(-3, 3), C(3, 5), D(1, 0) is the answer.

Step-by-step explanation:

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Forty families gathered for a fund-raising event. Suppose the individual contribution for each family is normally distributed with a mean and a standard deviation of $105 and $40, respectively. The organizers would call this event a success if the total contributions exceed $4,600. What is the probability that this fund-raising event is a success

Answers

Answer:

0.0571 = 5.71% probability that this fund-raising event is a success

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The organizers would call this event a success if the total contributions exceed $4,600

Forty families, so sample mean above 4600/40 = 115.

Mean and a standard deviation of $105 and $40

This means that [tex]\mu = 105, \sigma = 40[/tex]

Sample of 40

This means that [tex]n = 40, s = \frac{40}{\sqrt{40}} = 6.3246[/tex]

What is the probability that this fund-raising event is a success?

Sample mean above 115, which is 1 subtracted by the pvalue of Z when X = 115. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{115 - 105}{6.3246}[/tex]

[tex]Z = 1.58[/tex]

[tex]Z = 1.58[/tex] has a pvalue of 0.9429

1 - 0.9429 = 0.0571

0.0571 = 5.71% probability that this fund-raising event is a success

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Answers

the answer is C

there are 2 bars so you divide 2 by 1/8 to see how many eights there are

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