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2 years ago

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Andrew has a fish tank in the shape of a rectangular prism, and he needs to know the volume of water necessary to fill the tank.

He does not know the width of the tank, but he knows that the following graph models the volume of the tank, in cubic inches, when the tank has a width of w
what is the domain of the volume function?

A) (3, 5)

B) (0, ∞)

C) (0, 3)

D) (3, ∞)

2 answers:

Setler79 [48]2 years ago

7 0

Answer:

Option D is your Answer

Step-by-step explanation:

Alex2 years ago

3 0

Answer:

The correct option is D.

Step-by-step explanation:

Domain is the set of all posible

The volume of a rectangular prism is

$V=\text{Base area}\times height$

where,

$\text{Base area}=length\times width$

Dimensions can not be negative. So, the width and volume are always positive.

From the graph it is noticed that

In interval $(-\infty, -5)$ , width and volume is negative, so $(-\infty, -5)$ interval is not the domain.

In interval (-5,0), width is negative, so (-5,0) interval is not the domain.

In interval (0,3), volume is negative, so (0,3) interval is not the domain.

In interval $(3,\infty)$ , width and volume is negative, so $(3,\infty)$ interval is not the domain.

Therefore the correct option is D.

You might be interested in

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago

Which point on the graph represents the y-intercept? On a coordinate plane, a line goes through points V (negative 4, 4), W (0,

MatroZZZ [7]

Point W because the line intersects the y axis at W (0,2). sorry if it’s wrong :/

8 0

1 year ago

Read 2 more answers

The size of angle aob is equal to 132 degrees and the size of angle cod is equal to 141 degrees. find the size of angle dob.

Varvara68 [4.7K]

angle AOB = 132 and is also the sum of angles AOD and DOB. Hence
angle AOD + angle DOB = 132°    ---> 1

angle COD = 141 and is also the sum of angles COB and BOD. Hence
angle COB + angle DOB = 141°    ---> 2

Now we add the left sides together and the right sides of equations 1 and 2 together to form a new equation.

angle AOD + angle DOB + angle COB + angle DOB = 132 + 141       ---> 3

We should also note that:
angle AOD + angle DOB + angle COB = 180°

Therefore substituting angle AOD + angle DOB + angle COB in equation 3 by 180 and solving for angle DOB:
180 + angle DOB = 132 + 141
angle DOB = 273 - 180 = 93°

8 0

2 years ago

An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "

FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

$\large X_{sentences}$ the mean of the “sentences” group

$\large S_{sentences}$ the standard deviation of the “sentences” group

$\large X_{intentional}$ the mean of the “intentional” group

$\large S_{intentional}$ the standard deviation of the “intentional” group

Then, we can calculate by using the computer

$\large X_{sentences}=28.75$

$\large S_{sentences}=3.53553$

$\large X_{intentional}=31.625$

$\large S_{intentional}=1.40788$

$\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875$

The <em>standard error of the difference (of the means)</em> for a sample of size 8 is calculated with the formula

$\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}$

So, the standard error of the difference is

$\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546$

<em>In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%). </em>

<em>If this interval contains the zero, we can say there is no significant difference. </em>

<em>Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples. </em>

We get the $\large t_{0.05}$ which is a value of t such that the area under the Student's t distribution outside the interval $\large [-t_{0.05}, +t_{0.05}]$ is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

$\large t_{0.05}=2.365$

Now we can compute our confidence interval

$\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546$

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

6 0

2 years ago

Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies weig

Anon25 [30]

Answer:

3.84% probability that it has a low birth weight

Step-by-step explanation:

Problems of normally distributed samples are 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:

$Z = \frac{X - \mu}{\sigma}$

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3466, \sigma = 546$

If we randomly select a baby, what is the probability that it has a low birth weight?

This is the pvalue of Z when X = 2500. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2500 - 3466}{546}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384

3.84% probability that it has a low birth weight

3 0

2 years ago