For the equation -4y=8x what is the constant variation A. -4 B. -2 C. 1 D. 2

Simplify √ a^7 , where a>0 Which expression is equivalent to √a^7

Softa [21]

Answer:

$a^{3} \sqrt[]{a}$

Step-by-step explanation:

8 0

3 years ago

Round 3,989.23655 to the nearest thousandth

inysia [295]

3,989.237
A thousandth = 1/1000 = 0.001
Round 4th decimal place = 5
5 and over is rounding up and under 5 rounds down so it would be 3,989.237

6 0

3 years ago

A researcher is interested in determining if the more than two thirds of students would support making the Tuesday before Thanks

Yakvenalex [24]

Answer:

The null hypothesis is, {H₀}: p = 0.66, H₀: p = 0.66 .

The alternative hypothesis is, {Hₐ}: p > 0.66, Hₐ: p > 0.66

Step-by-step explanation:

Let p represent the proportion of students who are supportive of making the day before Thanksgiving a holiday.

Since two populations are considered, the alternative hypothesis for the difference of the means of the two populations has to be stated.

The proportion of two-third students are supportive of making the day before Thanksgiving a holiday is p = 0.66 obtained as shown below:

The null hypothesis is,

{H₀}:p = 0.66, H₀: p = 0.66

The alternative hypothesis is,

{Hₐ:p > 0.66, Ha: p > 0.66

The null hypothesis is, {H₀}: p = 0.66, H₀: p=0.66 .

The alternative hypothesis is, {Hₐ}: p > 0.66, Hₐ: p > 0.66

3 0

3 years ago

The cargo of the truck weighs less than 2,800 pounds. Use w to represent the weight (in pounds) of the cargo.

MakcuM [25]

Answer:

w < 2,800lb

Step-by-step explanation:

w weighs less than 2,800 pounds.

5 0

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.

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.

7 0

3 years ago