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3 years ago

The solution set of the equation is If √x + 3 = 3 -x is?

1 answer:

8 0

Two parallel strings of 12 and 20 cm are drawn in a circle. If the distance
between them it is 4 cm; How big is the diameter?

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Multiply3/4 x 5/8. Write in simplest form. Does the first factor<br> increase or decrease?

Zielflug [23.3K]

Answer:

-----

Step-by-step explanation:

3 5 15

-- * -- == ---

4 8 32

3 0

3 years ago

What is 3x + 10x =65?

zhannawk [14.2K]

Answer:

x = 5

Step-by-step explanation:

3x + 10x =65

13x = 65

/13 /13

------------------

x = 5

6 0

2 years ago

Read 2 more answers

Divide. (3 3/4 )÷(−2 1/2 ) Enter your answer as a mixed number, in simplified form, in the box.

Agata [3.3K]

Answer:

-11/14

Step-by-step explanation: first you divide 33/4 and (-21/2), then reduce the numbers to -33/4 times 2/21 after you reduce both fractions multiply -11/2 times 1/7 and you get -11/14 your alternative form would br -0.785714

8 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.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

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

What is the reciprocal of 2 1/3?

Oksi-84 [34.3K]

Answer:

2 3/1

Step-by-step explanation:

.................

6 0

3 years ago