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3 years ago

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A student scores 83, 90, and 85 on the first three science tests of the year in order to have a test average of at least 87, the

minimum score the student needs to get on the fourth test

1 answer:

Evgesh-ka [11]3 years ago

5 0

Hope you find this helpful!

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How do i find BC?<br><br> Khan Academy

Brrunno [24]

Answer:

using cos=adjacent/hypotenuse

Step-by-step explanation:

in this case cos(60)=x/4√3

x=cos(60)*4√3

x=3.46

6 0

2 years ago

HELP ME PLSSS ITS DUE TODAY

Gnom [1K]

Answer:

<u>S': (2, 1)</u>

<u>T': (5, 3)</u>

<u>U': (1, -4)</u>

<u>S'': (1, 3)</u>

<u>T'': (4, 5)</u>

<u>U'': (0, -2)</u>

Step-by-step explanation:

Hi!

For Reflection Across the X-Axis use this :

(x, y) -> (x, - y)

So :

S': (2, 1)

T': (5, 3)

U': (1, -4)

and then the question also asks for a translation so we follow what it gave us:

S'': (1, 3)

T'': (4, 5)

U'': (0, -2)

Please ask me any questions that you still may have!

and Have a great day! :)

5 0

8 months ago

Read 2 more answers

WILL GIVE BRAINLIEST 5 STARS AND THANKS Find the number of 4-digit numbers that contain exactly two even and two odd digits.

IgorLugansk [536]

Answer:

4433

Step-by-step explanation:

2 even 2 odd

4 0

2 years ago

Read 2 more answers

Serial numbers for a product are to be made using 2 letters followed by 3 numbers. If the letters are to be taken from the first

Daniel [21]

2 Letters chosen among 6 with no repeat:
a) the first has 6 ways
    the second has 5 ways

Then to choose 2 letters among 6, with no repeat, there are 6x5 = 30 ways

b) 3 numbers are taken from the digits 0 through 9 with repeats (total = 10)
   the first has 10 ways
   the second has 10 ways
    the third has 10 ways

Then to choose 3 numbers among 10, with repeat, there are 10x10x10= 1000 ways
And the total serial numbers that can be generated is
30 x 1000 = 3000

8 0

2 years ago

Some students are working together to draw a picture of a playground.

vodomira [7]

From the information provided by the statement, you know that

*Ann draws 2/16 of the playground

*Susan draws 6/16 of the playground because she draws 3 times as much as Ann so

$3\cdot\frac{2}{16}=\frac{3}{1}\cdot\frac{2}{16}=\frac{6}{16}$

*Louise draws 3/16 because she draws half as much as Susan, so

$\frac{1}{2}\cdot\frac{6}{16}=\frac{6}{32}=\frac{2\cdot3}{2\cdot16}=\frac{3}{16}$

*Sam draws 2/16 because she draws 1 less section than Louise

$\frac{3}{16}-\frac{1}{16}=\frac{2}{16}$

Now, let x be the fraction of the playground that still needs to be drawn. Then, you have

$\begin{gathered} \frac{2}{16}+\frac{6}{16}+\frac{3}{16}+\frac{2}{16}+x=\frac{16}{16} \\ \text{ Add similar terms} \\ \frac{13}{16}+x=\frac{16}{16} \\ \text{ Subtract }\frac{13}{16}\text{from both sides of the equation} \\ \frac{13}{16}-\frac{13}{16}+x=\frac{16}{16}-\frac{13}{16} \\ x=\frac{3}{16} \end{gathered}$

Therefore, 3/16 of the playground still needs to be drawn.

5 0

8 months ago