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m_a_m_a [10]
3 years ago
6

The ratio of Ds to As in the school was 4 to 72. If there were 1080 As in the school this term, how many Ds were there?

Mathematics
2 answers:
Vika [28.1K]3 years ago
3 0

Answer:

60

Step-by-step explanation:

1080/72 *15 = number if Ds

Ad libitum [116K]3 years ago
3 0

Answer:

60 D's

Step-by-step explanation:

We are given that the ratio of D to A is 4 to 72,

We can write that as 4:72.

Since students got 1080 A's. we can divide 1080 by 72 to help us find the answer.

1080/72=15

We know that the relationship between the amount of D's simplified and the actual amount of D's is equal to 15 times. So we do the given amount (4) multiplied by 15 (4*15) to get 60.

Please let me know if you have any questions.

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sashaice [31]

Answer:

If its a line withe 8x as one of the factors in the mx position of y=mx+b then its parrallel

Step-by-step explanation: Hopeit helped

For example y=8x+2y is parallel to y=8x+3y

5 0
3 years ago
Can someone check my answer? i got 600.
lubasha [3.4K]

Answer:

Volume = 120 cubic cm

Step-by-step explanation:

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7 0
3 years ago
HELP ASAP
Alex787 [66]

<span>v(x)=(s/t)
= (3x - 6) / (-3x+6)
= [3(x-2)] / [-3(x-2)] --> 3 is factored out
= 1/-1 </span>---> common terms are cancelled out. 
= -1 ---> This is the simplified formula.

To find the domain, we equate the denominator to 0.
-3x+6 = 0
3x = 6
x = 2
Domain: all values except 2.

w(x)=(t/s)(x)
= (-3x+6)x / (3x-6)
= [-3x(x-2)] / [3(x-2)] --> 3 is factored out
= -x --> The common terms are cancelled out. This is the simplified formula.

Solving for domain:
3x-6 = 0
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x = 2
Domain: all values except 2.

4 0
3 years ago
The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas
Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is = \frac{(n+1)}{2}.

\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5  position orders

10^{th} \ and \ 11^{th} place an average of observations so, the

Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

Q_1 often falls throughout the ordered BELOW average is = \frac{(n+1)}{2} place.

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Average 5^{th} \ and \ 6^{th}  location findings

Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5

In  = \frac{(n+1)}{2} the ordered place, Q3 always falls ABOVE the median.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th} ordered At median ABOVE 

Consequently Q_3 falls between 5^{th} \ and \ 6^{th} ABOVE the median position

15^{th}\  and \ 16^{th}place average of observations

\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2}  = 502

\to IQR = Q_3 - Q_1= 502-358.5= 143.5  

\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25

4 0
3 years ago
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