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3 years ago

(05.03)The equation shows the relationship between x and y:

Mathematics

2 answers:

Aleks04 [339]3 years ago

6 0

Slope = -2
Hope that helps
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Aleks04 [339]3 years ago

3 0

The slope of this line is the number that is attached to the x...so in this case it would be -2 :) hope this helps you out! good luck!

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3. Two of the vertices of an equilateral triangle are (0, -a)

egoroff_w [7]

The third coordinate of the equilateral triangle is $(\sqrt3a,0)$ . Ir can be obtained by distance formula.

What is distance formula?

It gives the distance between two coordinates.

Distance formula is given by the following expression:

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Consider the third cordinate is (x,0) and x is positive because it is on tight side on the x axis.

Now, since the triangle is equilateral hence, the distance of (x,0) from (0, -a) and (0, a) should be equal to distance between (0, -a) and (0, a).

Now, caculate the distance between (0,-a) and (0, a). $\sqrt{(0-0)^2+(-a-a)^2}=2a$

Now, distance between (x,0) and (0, a) should be 2a.

$\sqrt{(x-0)^2+(0-a)^2}=2a\\\sqrt{x^2+a^2}=2a$

Now, square on both the sides.

$x^2+a^2=4a^2\\x^2=3a^2\\x=\sqrt3a$ and $-\sqrt3a$

But the negative value of x is not possible because the coordinate lies on the right side.

Hence, the third coordinate of equilateral triangle is $(\sqrt3a,0)$ .

Learn more about distance formula from the following link:

brainly.com/question/7243416

#SPJ1

5 0

1 year ago

Simplify 2⁰+5¹+4³/7

Alekssandra [29.7K]

Your question has been heard loud and clear.

2^0= 1

5^1= 5

4^3=64

2^0+5^1+4^3/7= 1+5+64/7 = 6+64/7= 15.14

So , 2^0+5^1+4^3/7 = 15.14

Thank you

8 0

4 years ago

A deer population is declining by 2.2% per year. The population can be modelled using the formula p = 240(0.978)n , where P is t

ale4655 [162]

Answer:

The current deer population is 240.

Step-by-step explanation:

The expression given models a exponential decay follows:

$p(n) = 240*(0.978)^n$

For every year that passes since the beginning of the count the population is 0.978 of the year prior. Assuming that the count starts at this year, then "n" is equal to 0, applying this value of n to the formula gives us:

$p(0) = 240*(0.978)^0\\p(0) = 240*1 = 240$

The current deer population is 240.

3 0

3 years ago

At a physics convention, 10 companies set up equal sized square booths in a row along one wall of the convention center. The boo

exis [7]

I'm not 100% sure what this all means because I'm in a different school but, I believe they have a side length of 4600.

P.S. If it doesn't work, I am so sorry

3 0

3 years ago

Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak

crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

8 0

3 years ago