All categories

3 years ago

I tried graphing this and I messed up can someone help me graph it?

Mathematics

1 answer:

IrinaVladis [17]3 years ago

7 0

For x=-y+3, you add y to both sides which gives you x+y=3 then you subtract x from both sides which gives you y=-x+3.

You might be interested in

PLEASE PLEASE HELPPPPPP

murzikaleks [220]

Answer:

Step-by-step explanation:

The rulers tell you the length of the sides

8 0

2 years ago

$5 PAYPAL + BRAINLIEST TO ANSWER THS

expeople1 [14]

Answer:

Quadratic Equation:

$3x^2=2x +5$

$\text{Standard Form: } 3x^2-2x-5=0$

From the standard form of a Quadratic Function, we get:

$a=3,\:b=-2,\:c=-5$

Discriminant:

$\Delta=\left(-2\right)^2-4\cdot \:3\left(-5\right)$

$\Delta=\left(-2\right)^2+4\cdot \:3\cdot \:5$

$\Delta=64$

From the discriminant, we conclude that the equation will have two real solutions.

State that:

$b^2-4ac$

$b^2-4ac =0:\text{The equation has 1 real solution}$

$b^2-4ac >0:\text{The equation has 2 real solutions}$

By the way, solving the equation given:

$x=\frac{2\pm\sqrt{64}}{2\cdot \:3}$

$x=\frac{2\pm\sqrt{64}}{6}$

$x=\frac{2\pm8}{6}$

$x_{1} =\frac{10}{6}=\frac{5}{3}$

$x_{2}=\frac{-6}{6} =-1$

5 0

3 years ago

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon

fiasKO [112]

Answer:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.

This means that $n = 603, \pi = \frac{142}{603} = 0.2355$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694$

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

8 0

2 years ago

What number is 62% of 50

goldfiish [28.3K]

$62\% \ of \ 50= 62\% \times 50=\frac{62}{100} \times 50= \frac{62}{2} \times 1=31$

31 is 62% of 50.

8 0

2 years ago

In old-growth forests of douglas fir, the spotted owl dines mainly on flying squirrels. If we left When the initial population i

Delicious77 [7]

Answer: There are 667 owls for every 1000 squirrels.

Step-by-step explanation:

Since we have given that

Number of owls = 30

Number of squirrels = 45

If the number of owls and squirrels will continue to increase.

If there are 1000 squirrels, then the number of owls would be

$\dfrac{30}{45}=\dfrac{x}{1000}\\\\x=\dfrac{30\times 1000}{45}\\\\x=666.67\\\\x\approx 667$

Hence, there are 667 owls for every 1000 squirrels.

5 0

2 years ago