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2 years ago

8

Fort-eight students signed up to volunteer for a race. There were 3 equal groups of students and each group had a different task

. How many students were in each group?

1 answer:

notka56 [123]2 years ago

4 0

Answer:

16

Step-by-step explanation:

48 ÷ 3 = 16

hope this helps...

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Without graphing, classify each system of equations as independent, dependent, or inconsistent. Solve independent systems by gra

MArishka [77]

1. Is Dependent.
2. Is Inconsistent

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2 years ago

If f(x)=x^2-5 and g(x)=6x then g(f(x)) is equal to

Romashka-Z-Leto [24]

G(x) = 6x

<span>g(f(x)) </span>
<span>= 6((x^2) - 5) </span>
<span>= (6x^2) - 30
</span>
so its equal to
<span>(6x^2) - 30</span>

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2 years ago

Read 2 more answers

X=3y + 1<br> 2x + 4y =12

Alika [10]

Answer:

the solution is (4, 1).

Step-by-step explanation:

The goal here is to "solve this system of linear equations." You need to share that along with the equations.

We can begin the solution by substituting 3y + 1 for x in the second equation:

2(3y + 1) + 4y = 12

Performing the multiplication, we get 6y + 2 + 4y = 12, or 10y = 10.

Then y must be 1.

Substituting 1 for y in the first equation, we get x = 3(1) + 1 = 4

Thus, the solution is (4, 1).

4 0

2 years ago

Please help ASAP! Thanks! For the figures below, assume they are made of semicircles, quarter circles and squares. For each shap

bazaltina [42]

Answer:

Part 1) $A=36(\pi-2)\ cm^2$

Part 2) $P=6(\pi+2\sqrt{2})\ cm$

Step-by-step explanation:

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

so

$A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)$

we have that the base and the height of triangle is equal to the radius of the circle

$r=12\ cm\\b=12\ cm\\h=12\ cm$

substitute

$A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)$

$A=(36\pi-72)\ cm^2$

simplify

Factor 36

$A=36(\pi-2)\ cm^2$

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

<em>The circumference of a quarter of circle is equal to</em>

$C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r$

substitute the given values

$C=\frac{1}{2}\pi (12)$

$C=6\pi\ cm$

The hypotenuse of right triangle is equal to (applying the Pythagorean Theorem)

$AC=\sqrt{12^2+12^2}$

$AC=\sqrt{288}\ cm$

simplify

$AC=12\sqrt{2}\ cm$

Find the perimeter

$P=(6\pi+12\sqrt{2})\ cm$

simplify

Factor 6

$P=6(\pi+2\sqrt{2})\ cm$

6 0

2 years ago

Three machines A, B, and C produce 55%, 25% and 20% respectively. The percentages of defective output are 3.5%, 4.5% and 5.5%. I

pishuonlain [190]

We are given the following information concerning the three production machines;

$\begin{gathered} Of\text{ the total production;} \\ A=55\text{ \%}=0.55 \\ B=25\text{ \%}=0.25 \\ C=20\text{ \%}=0.20 \end{gathered}$

Also, we are given the percentage of defective output as follows;

$\begin{gathered} A=3.5\text{ \%}=0.035 \\ B=4.5\text{ \%}=0.045 \\ C=5.5\text{ \%}=0.055 \end{gathered}$

Therefore, if an item is selected randomly, the probability that the item is defective would be;

$\begin{gathered} P\lbrack defective\rbrack=(0.55\times0.035)+(0.25\times0.045)+(0.20\times0.055) \\ P\lbrack\text{defective\rbrack}=0.01925+0.01125+0.011 \\ P\lbrack\text{defective\rbrack}=0.0415 \end{gathered}$

ANSWER:

The probability that the item is defective would be 0.0415

6 0

9 months ago