Solve each system by ELIMINATION<br> 1) -5x-Y=21<br>

arlik [135]

$S\frac{O}{H} C\frac{A}{H} T\frac{O}{A}$ or $Sin\frac{Opposite}{Hypotenuse}Cos\frac{Adjacent}{Hypotenuse} Tan\frac{Opposite}{Adjacent}$

At the point/angle E:

- the adjacent side(side <u>next to</u> the point/angle) is ED

- the opposite side(side of the triangle <u>across</u> from the point/angle) is DF

- the hypotenuse (the<u> longest side</u> of the triangle) is EF.

sin ∠E = $\frac{opposite}{hypotenuse}$

sin ∠E = $\frac{6}{10} = \frac{3}{5}$ (simplified)

5 0

3 years ago

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The midpoint of UV is (5, -10). The coordinates of one endpoint are U(3,6). Find the coordinates of endpoint V.

Grace [21]

The answer is (7, -26) for The second endpoint.

We'll call the midpoint M. In order to find this, we must first note that to find a midpoint we need to take the average of the endpoints. To do this we add them together and then divide by 2. So, using that, we can write a formula and solve for each part of the k coordinates. We'll start with just x values.

(Ux + Vx)/2 = Mx

(Vx + 3)/2 = 5

Vx + 3 = 10

Vx = 7

And now we do the same thing for y values

(Uy + Vy)/2 = My

(Vy + 6)/2 = -10

Vy + 6 = -20

Vy = -26

This gives us the final point of (7, -26)

6 0

2 years ago

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54 out of the 60 magazines at the bookstore are women's magazines. What percentage of the magazines at the bookstore are women's

Nataly_w [17]

Answer:

90

Step-by-step explanation:

for percentage

$\frac{54}{60}$ ×100

= 0.9×100

=90

6 0

3 years ago

PLS HELP, HOMEWORK. (15 points)

Rus_ich [418]

There is no problem on here

5 0

2 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>

6 0

3 years ago

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Solve each system by ELIMINATION 1) -5x-Y=21 - 4x Y=6