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algol [13]
3 years ago
11

The circle graph shows how Tremaine divided his time on the computer last week. Tremaine used the computer a total of 30 hours l

ast week. How many more hours did Tremaine use the computer to play games than to do research?
A) 20 hours
B) 6 hours
C) 1.5 hours
D) 7.5 hours

Mathematics
2 answers:
Lynna [10]3 years ago
7 0
25-5=20 so the answer is A) 20 hours
Tomtit [17]3 years ago
6 0

Answer:

A) 20 hours

Step-by-step explanation:

:>>>

25-5=20

You might be interested in
Which ordered pair would form a proportional relationship with the point graphed below? (60,-20)
valentina_108 [34]

Given:

The graphed point is (60,-20).

To find:

The ordered pair that would form a proportional relationship with the given point.

Solution:

If y is proportional to x, then

y\propto x

y=kx

\dfrac{y}{x}=k

Where, k is the constant of proportionality.

For the given point,

k=\dfrac{-20}{60}

k=\dfrac{-1}{3}

For option (A),

k=\dfrac{30}{-10}

k=-3\neq \dfrac{-1}{3}

For option (B),

k=\dfrac{-15}{30}

k=-\dfrac{1}{2}\neq \dfrac{-1}{3}

For option (C),

k=\dfrac{10}{-30}

k=\dfrac{-1}{3}.

The point (-30,10) gives the same value of the constant of proportionality. So, the point (-30,10) forms a proportional relationship with the given point.

Therefore, the correct option is C.

7 0
2 years ago
Please help me!!
Tomtit [17]

the area would be 4 times more

 so 135 x 4 = 540

3 0
3 years ago
Simplify the expression (x)^3(x^3y)^2
larisa86 [58]

Answer:

A)

Step-by-step explanation:

(a^{m})^{n} = a^{mn}\\\\x^{3}*(x^{3}*y^{2})^{2}=x^{3}*(x^{3})^{2}*(y)^{2}\\\\\\=x^{3}*x^{3*2}*y^{2}\\\\\\=x^{3}*x^{6}*y{2}\\\\\\=x^{3+6}*y^{2}\\=x^{9}y^{2}

6 0
3 years ago
Solve pls brainliest
Murrr4er [49]

Answer:

-1, -24

Step-by-step explanation:

-9-(-8)

double negative = positive

-9+8=-1

21-45=-24

6 0
2 years ago
PLEASE HELP
dsp73

Answer:

All events are independent

Step-by-step explanation:

You are given the table

\begin{array}{cccc}&\text{Chocolate}&\text{Vanilla}&\text{Total}\\\text{Adults}&0.21&0.39&0.60\\\text{Children}&0.14&0.26&0.40\\\text{Total}&0.35&0.65&1.00\end{array}

Two events A and B are independent when

Pr(A\cap B)=Pr(A)\cdot Pr(B)

a) A="Chocolate"

B="Adults"

A and B="Chocolate and Adults"

Pr(A)=0.35\\ \\Pr(B)=0.60\\ \\Pr(A\cap B)=0.21

Since 0.35\cdot 0.60=0.21 events are independent

b) A="Children"

B="Chocolate"

A and B="Children and Chocolate"

Pr(A)=0.40\\ \\Pr(B)=0.35\\ \\Pr(A\cap B)=0.14

Since 0.40\cdot 0.35=0.14 events are independent

c) A="Vanilla"

B="Children"

A and B="Vanilla and Children"

Pr(A)=0.65\\ \\Pr(B)=0.40\\ \\Pr(A\cap B)=0.26

Since 0.65\cdot 0.40=0.26 events are independent

4 0
3 years ago
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