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3 years ago

Indicate the equation of the given line in standard form.

Mathematics

2 answers:

ASHA 777 [7]3 years ago

7 0

You must convert point slope form to standard form in this problem. point= (h,k)

POINT SLOPE:
y-(-1)=(3/4)(x-2)
y+1=(3/4)x-(3/2)
y= (3/4)x-(5/2)

Andrei [34K]3 years ago

5 0

Answer:

3x-4y=10

Step-by-step explanation:

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If f(x) = 3x + 2 and g(x) = x2<br> x, find each value.<br> 18. f(2)=

Pavel [41]

F(2) = 8
if x = 2 then you plug in 2 for x making the equation 3(2) + 2 which equals 8

7 0

3 years ago

30% of Number A is the same as 40% of Number B.

Goryan [66]

30% of number A means

$\dfrac{30A}{100}=\dfrac{3A}{10}$

similarly, 40% of number B means

$\dfrac{40B}{100}=\dfrac{2B}{5}$

These must be equal, so we have

$\dfrac{3A}{10}=\dfrac{2B}{5}$

Multiply both sides by 10 to get

$3A=4B \iff A=\dfrac{4B}{3}$

So, you can choose any number for A, and then you multiply it by 4/3 to get B. There are infinitely many solutions to this equation.

5 0

3 years ago

One Morning in Atlanta, Tina observed that the temperature was 72 degrees. Later the temperature rose 21 degrees and then droppe

lesya [120]

answer: 72+21=93-11=82

5 0

3 years ago

☺

Fittoniya [83]

Answer: 1 teashirt=7 L

1 hat= 3L

Step-by-step explanation:

2 T shirts= 31 - 17

2 T shirts= 14

1 T shirt=14/2

1 T = 7 L

14L+ 1H=17 L

one hat = 17 - 14=3 L

8 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