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3 years ago

Which of the following equations have infinitely many solutions?

Mathematics

1 answer:

Svetllana [295]3 years ago

3 0

Answer:

Choice A

Step-by-step explanation:

In each case, each equation has an equation of a line in y = mx + b form equaling another equation of a line in y = mx + b form. If the two sides are equal, it is the same equation, there are infinitely many solutions. If the sides are different, then if the slopes are different, the lines intersect at one point, and there is exactly 1 solution. If the slopes are equal, the lines are parallel, and there is no solution.

(Choice A) -10x-10=-10x-10

In Choice A, both sides of the equation are equal, so there are infinitely many solutions.

(Choice B) 10x-10=-10x+10

(Choice C) 10x-10=-10x-10

(Choice D) -10x-10=-10x+10

In choices B through D, the two sides are not equal, so there is either 1 solution (B and C since they have different slopes) or no solution (D since the slopes are equal).

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A quadrilateral WXYZ has vertices W(3, −5), X(1, −3), Y(−1, −5), and Z(1,−7). What are the vertices of r(90, O)(WXYZ)?

VashaNatasha [74]

Given:

A quadrilateral WXYZ has vertices W(3, −5), X(1, −3), Y(−1, −5), and Z(1,−7).

Rule of rotation is $r_{(90^\circ, O)}(WXYZ)$ .

To find:

The vertices after rotation.

Solution:

We know that, $r_{(90^\circ, O)}(WXYZ)$ means 90 degrees counterclockwise rotation around the origin.

So, the rule of rotation is defined as

$(x,y)\to (-y,x)$

Using this rule, we get

$W(3,-5)\to W'(5,3)$

$X(1,-3)\to X'(3,1)$

$Y(-1,-5)\to Y'(5,-1)$

$Z(1,-7)\to Z'(7,1)$

Therefore, the required vertices after rotation are W'(5,3), X'(3,1),Y'(5,-1) and Z'(7,1).

8 0

3 years ago

ASAP PLZZZ I NEED HELP

son4ous [18]

Answer:

994.5

Step-by-step explanation:

First, you need to figure out how many people are actually attending the event. It says that there are 50% more so we are going to have to add 56% of 150 to 150.

150 x 0.56 = 84

150 + 84 = 234

Now, we have to multiply this by 4.25, because this is the cost of the food per person.

234 x 4.25 = 994.5

3 0

3 years ago

So how do i so how do I solve to get this answer

Elina [12.6K]

Step-by-step explanation:

First, You would solve what you can which would be the equation on the left Which will get you 4096. After that its just solving for P

4096=(4^p)^2

64=4^p (you would take the square root of 4096 to get 64)

Then, Since 64 is a perfect Cube root of 4, P will equal 3

(Please award this Brainliest and Thank the skeleton in your closet)

8 0

3 years ago

Find the circumference and area

KATRIN_1 [288]

18.85 I hope this helps you :)

5 0

3 years ago

Read 2 more answers

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.

7 0

3 years ago