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4 years ago

Which equation represents a proportional relationship?

Mathematics

2 answers:

bonufazy [111]4 years ago

4 0

It's the first one. y = 3x
proportional means they are just some multiple of the other.

Klio2033 [76]4 years ago

3 0

Y=3x hope this helps

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Given f(x) = -4x - 2, solve for x when f(x) = 6.

Andrej [43]

Answer:

use

Step-by-step explanation:

calculator

6 0

3 years ago

Trade associations, such as the United Dairy Farmers Association, frequently conduct surveys to identify characteristics of thei

asambeis [7]

Complete Question

Trade associations, such as the United Dairy Farmers Association, frequently conduct surveys to identify characteristics of their membership. If this organization conducted a survey to estimate the annual per-capital consumption of milk and wanted to be 95% confident that the estimate was no more than 0.5 gallon away from the actual average, what sample size is needed? Past data have indicated that the standard deviation of consumption of approximately 10 gallons.

Answer:

The sample size is $n = 1537 \ gallons$

Step-by-step explanation:

From the question we are told that

The margin of error is $MOE = 0.5$

The confidence level is $C = 95$ %

Given that the confidence level is 95% the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5$ %

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table , the values is $Z_{\frac{\alpha }{2} } = 1. 96$

The reason we are obtaining critical values of

$\frac{\alpha }{2}$

instead of

$\alpha$

is because

$\alpha$

represents the area under the normal curve where the confidence level interval (

$1-\alpha$

) did not cover which include both the left and right tail while

$\frac{\alpha }{2}$

is just the area of one tail which what we required to calculate the sample size

Now the sample size is mathematically represented as

$n = \frac{[Z_{\frac{\alpha }{2} }] ^2 * \sigma ^2}{MOE^2}$

substituting values

$n = \frac{1.96^2 * 10 ^2}{0.5^2}$

$n = 1537 \ gallons$

7 0

3 years ago

A double cheeseburger has 540 calories, a milk shake has 490 calories and a serving of fries has 380 calories. If you burn 130 c

Inessa05 [86]

Answer:

10.846 hours

Step-by-step explanation:

first you add all the calories= 1410 calories

then you divide 1410 calories by 130 (the calories you can burn in one hour)

your answer is 10.846 which means that it takes 10.846 hours to consume 1410 calories.

4 0

3 years ago

A large building has 1000 windows, and 5 of the windows need to be replaced. what decimal represents the number of Windows that

Taya2010 [7]

The answer is .005 because five out of one thousand need to be replaced.

4 0

3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

4 years ago