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2 years ago

The graph provided show the volume (in cubic inches) of air in a balloon as it chang

Mathematics

1 answer:

drek231 [11]2 years ago

5 0

The volume of the balloon is 500 cubic inches after 20 seconds

<h3>How to determine the volume?</h3>

From the given graph, we have:

Time on the x-axis
Volume on the y-axis

Also from the graph, we have:

V(20) = 500

This means that:

The volume of the balloon is 500 cubic inches after 20 seconds

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What is the location of the point (5, 0) translates 4 units to the down and reflected across the y-axis?

CaHeK987 [17]

STEP-BY-STEP EXPLANATION:

Given information

The given ordered point = (5, 0)

Step 1: We need to translate the point 4 units down

To translate down means we will be subtracting a value from the y--axis

Hence, we have

$\begin{gathered} (x,\text{ y) }\rightarrow\text{ (x, y-b)} \\ \text{where b = 4} \\ (5,\text{ 0) }\rightarrow\text{ (5, 0 - 4)} \\ (5,\text{ 0) }\rightarrow\text{ (5, -4)} \end{gathered}$

When translated 4 units down, we got (5, -4)

Step 2: Reflect over the y-axis

The general rule for reflecting over the y-axis is (-x, y)

This means the value of x will be negated and the value of y will remain the same

$\begin{gathered} \text{Over the y-ax}is \\ (x,\text{ y) }\rightarrow\text{ (-x, y)} \\ (5,\text{ -4) }\rightarrow\text{ (-5, -4)} \end{gathered}$

Step 3: the graph the point

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1 year ago

Express each ratio as a unit rate. 325.4 miles on 17.3 gallons having trouble :/

kkurt [141]

Hello! I can help you with this question! To find the unit rate, we’ll have to divide total by number of units. Or in this case, miles/gallons. 325.4/17.3 is 18.809 and other numbers, or 18.81 when rounded to the nearest hundredth. If you’re looking for the answer to the nearest hundredth, the answer is 18.81 mpg. Or, if you’re looking for the answer to the nearest whole number, the answer is 19 mpg.

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3 years ago

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elixir [45]

Answer:

Step-by-step explanation:

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2 years ago

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A baby grows 4 inches every 6 months. What is his rate of change?

ser-zykov [4K]

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4:6 but if you reduce it is 2:3

Step-by-step explanation:

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3 years ago

A county environmental agency suspects that the fish in a particular polluted lake have elevated mercury levels. To confirm that

suter [353]

Answer:

a. The 95% confidence interval for the difference between means is (0.071, 0.389).

b. There is enough evidence to support the claim that the fish in this particular polluted lake have signficantly elevated mercury levels.

c. They agree. Both conclude that the levels of mercury are significnatly higher compared to a unpolluted lake.

In the case of the confidence interval, we reach this conclusion because the lower bound is greater than 0. This indicates that, with more than 95% confidence, we can tell that the difference in mercury levels is positive.

In the case of the hypothesis test, we conclude that because the P-value indicates there is a little chance we get that samples if there is no significant difference between the mercury levels. This indicates that the values of mercury in the polluted lake are significantly higher than the unpolluted lake.

Step-by-step explanation:

The table with the data is:

Sample 1 Sample 2

0.580 0.382

0.711 0.276

0.571 0.570

0.666 0.366

0.598

The mean and standard deviation for sample 1 are:

$M=\dfrac{1}{5}\sum_{i=1}^{5}(0.58+0.711+0.571+0.666+0.598)\\\\\\ M=\dfrac{3.126}{5}=0.63$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{5}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(0.58-(0.63))^2+...+(0.598-(0.63))^2]}\\\\\\ s=\sqrt{\dfrac{1}{4}\cdot [(0.002)+(0.007)+(0.003)+(0.002)+(0.001)]}\\\\\\ s=\sqrt{\dfrac{0.015}{4}}=\sqrt{0.0037}\\\\\\s=0.061$

The mean and standard deviation for sample 2 are:

$M=\dfrac{1}{4}\sum_{i=1}^{4}(0.382+0.276+0.57+0.366)\\\\\\ M=\dfrac{1.594}{4}=0.4$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{4}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{3}\cdot [(0.382-(0.4))^2+(0.276-(0.4))^2+(0.57-(0.4))^2+(0.366-(0.4))^2]}\\\\\\ s=\sqrt{\dfrac{1}{3}\cdot [(0)+(0.015)+(0.029)+(0.001)]}\\\\\\ s=\sqrt{\dfrac{0.046}{3}}=\sqrt{0.015}\\\\\\s=0.123$

Confidence interval

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=5 has a mean of 0.63 and a standard deviation of 0.061.

The sample 2, of size n2=4 has a mean of 0.4 and a standard deviation of 0.123.

The difference between sample means is Md=0.23.

$M_d=M_1-M_2=0.63-0.4=0.23$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{0.061^2}{5}+\dfrac{0.123^2}{4}}\\\\\\s_{M_d}=\sqrt{0.001+0.004}=\sqrt{0.005}=0.07$

The critical t-value for a 95% confidence interval is t=2.365.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.365 \cdot 0.07=0.159$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 0.23-0.159=0.071\\\\UL=M_d+t \cdot s_{M_d} = 0.23+0.159=0.389$

The 95% confidence interval for the difference between means is (0.071, 0.389).

Hypothesis test

This is a hypothesis test for the difference between populations means.

The claim is that the fish in this particular polluted lake have signficantly elevated mercury levels.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0$

The significance level is 0.05.

The sample 1, of size n1=5 has a mean of 0.63 and a standard deviation of 0.061.

The sample 2, of size n2=4 has a mean of 0.4 and a standard deviation of 0.123.

The difference between sample means is Md=0.23.

$M_d=M_1-M_2=0.63-0.4=0.23$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{0.061^2}{5}+\dfrac{0.123^2}{4}}\\\\\\s_{M_d}=\sqrt{0.001+0.004}=\sqrt{0.005}=0.07$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.23-0}{0.07}=\dfrac{0.23}{0.07}=3.42$

The degrees of freedom for this test are:

$df=n_1+n_2-1=5+4-2=7$

This test is a right-tailed test, with 7 degrees of freedom and t=3.42, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t>3.42)=0.006$

As the P-value (0.006) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the fish in this particular polluted lake have signficantly elevated mercury levels.

c. They agree. Both conclude that the levels of mercury are significnatly higher compared to a unpolluted lake.

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3 years ago