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igomit [66]
3 years ago
9

Your digital camera has a 512 megabyte memory card. You take pictures at two resolutions, a low resolution requiring 4 megabytes

of memory per photo and a high resolution requiring 8 megabytes of memory per photo. Write an equation to model the amount of each kind of photo you can take.
Mathematics
1 answer:
____ [38]3 years ago
6 0

One possible answer is 8H + 4L = 512

=======================================================

Explanation:

  • H = number of high resolution photos
  • L = number of low resolution photos

H and L are placeholders for positive whole numbers, or 0 could replace either variable.

1 high resolution photo takes up 8 mb of space, so H of them take up 8H megabytes of space.

1 low resolution photo takes up 4 mb of space, so L of them take up 4L megabytes.

Overall, the two types of photos take up 8H+4L megabytes, which is equal to 512 since that's the max capacity. That leads to the equation 8H+4L = 512

----------------------------------

Extra info:

There are other ways to express the equation 8H+4L = 512. We could divide each part by 4 to get 2H+L = 128, but I think this equation loses its descriptive quality in a way. We no longer can see that each high resolution photo takes up 8 megabytes (instead we might mistakenly think only 2 megabytes are used per high resolution photo). A similar mistake may happen with the low resolution photos also.

The equation 2H+L = 128 can be rearranged into L = -2H+128 when solving for L.

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Find the radius of a circle using the Pythagorean theorem, given that the center is at (1, 5) and the point (5, 1) lies on the c
Assoli18 [71]

Given :

Center of circle, C = ( 1 , 5 )

Point on circle, P = ( 5 , 1 )

To Find :

The radius of a circle.

Solution :

We know, radius of circle is the distance between center and point of circle.

Distance between two points A(p,q) and B(r,s) is given by :

D = \sqrt{(p-r)^2+(q-s)^2}

So, distance between the points (1, 5) and (5, 1) is :

D = \sqrt{(1-5)^2+(5-1)^2}\\\\D = 5.66 \ units

Therefore, radius of circle is 5.66 units.

Hence, this is the required solution.

4 0
3 years ago
 Triangle PQR has two known interior angles of 24°, and 100°.
Tamiku [17]
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3 0
3 years ago
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If f(x)=x2 and g(x)=|x| find the following functions. 1. f+g 2. f-g 3. fg​
marusya05 [52]

Answer:

<u>1. x^2 + (absolute value) X</u>

<u>2. x^2 - (absolute value) X</u>

<u>3. x^2 * (absolute value) X</u>

Step-by-step explanation:

Add equations for 1.

Subtract equations for 2.

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5 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

<u>Confidence interval</u>

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).

<u>Hypothesis test</u>

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.

<u> </u>

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
find the equation for the line that passes through the point (5, 2), and that is parallel to the line with the equation x=-4
BabaBlast [244]
Y-y1=m(x-x1)
y-2=-4(x-5)
y-2=-4x+20
y-2+2=-4x+20+2
y=-4x+22
6 0
3 years ago
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