Please help me quickly!

Jessica is 5 years older than her sister Jenna. Jenna tells her sister that in 5 years, she will be as old as Jessica was 5 year

vlabodo [156]

Answer:

x + 5 = (x + 5) - 5, which has no solution ⇒ answer C

Step-by-step explanation:

* Lets study the situation in the problem

- Jessica is 5 years older than her sister Jenna

- After five years Jenna's age will be as old as Jessica was five years ago

- Jenna's age now is x

* Lets change all information above to equation

∵ Janna's age now is x

∵ Jessica is 5 years old than Janna

∴ The age of Jessica now is x + 5

- After 5 years Janna's age will be add by 5 years

∵ Her age now is x

∴ Her age after 5 years will be x + 5

- From 5 years ago Jessica's age was her age now mins 5 years

∵ Her age now is x + 5

∴ Her age from 5 years ago is (x + 5) - 5

∵ Janna's age after 5 years = Jessica age from 5 years ago

∴ x + 5 = (x + 5) - 5

- Lets solve the equation to know how many solution

∵ x + 5 = x + 5 - 5 ⇒ add the like terms

∴ x + 5 = x ⇒ subtract x from both sides

∴ 5 = 0

- But 5 ≠ 0, the two sides of the equation not equal each other

∴ There is no solution for this equation

* The equation is x + 5 = (x + 5) - 5, which has no solution

8 0

3 years ago

Read 2 more answers

A gutter at the edge of a roof drops 2 inches for every 30 feet of length. The measure of the angle to the nearest tenth of a de

g100num [7]

About 0.3 degrees, by estimation.
The sixty-to-one rule is useful here. At a distance of 60 units, the angle in degrees and the distance (in units) are about equal for small angles.
Thirty to two inches is the same as sixty to four inches, about a third of a foot, so the angle must be about a third of a degree. Rounded it gives 0.3.
My horribly antiquated TI-82 thinks the answer is about 0.3183 by this methodtan−1(1/180)≈0.3183

8 0

2 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

A number with 2 or more factors.

Molodets [167]

Answer:

composite number. that is what it is called

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the surface area of the prism or pyramid. Round to the nearest hundredth if necessary.

frutty [35]

Answer:

the total surface area would be 233^2 im pretty sure (im not 100% sure) because the you do the lateral surface area add the top surface area add the bottom surface area so im pretty sure the answer would be 233cm^2 (^2 means squared)

8 0

3 years ago