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3 years ago

I’ll give u a kiss if u help me with this ASAP PLZZZ;) with an explanation too plzzz

Mathematics

1 answer:

Dvinal [7]3 years ago

8 0

Can i get a kiss??? x#*||~~

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In a triangle ABC, tan (a) = 20/21. Find sin (a) and cos (a)

sesenic [268]

Answer:

sin(a)=20/29

cos(a)=21/29

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3 years ago

Ms evans paycheck is $850 a week.If she works 49 hours per week, how muxh does she make an hour

Aneli [31]

She would make about 17.35 because $850 divided by 49 is 17.3469388 rounded is 17.35

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2 years ago

45 + 60 = 15 x ___ + 15 x ___

tia_tia [17]

Answer:45+60=15x+15x

Step 1: Simplify both sides of the equation.

45+60=15x+15x

(45+60)=(15x+15x)(Combine Like Terms)

105=30x

Step 2: Flip the equation.

30x=105

Step 3: Divide both sides by 30.

30x/30= 105/30

x= 7/2

Answer:

x= 7/2

may i have brainliest please

3 0

3 years ago

Simplify the expression 417000 x 0.0002

Amanda [17]

Please see picture above.

5 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

2 years ago