Evaluate 7 + (-3X2) for x = 0.

PLEASE HELP !!! The height of a rectangle is twice the width and the area is 32. Find the dimensions.

Helen [10]

Answer:

The rectangle has a width of 4 and a height of 8

Step-by-step explanation:

Let the height of the rectangle be H and the width be W.

We know the height of the rectangle is twice the width, so:

H = 2W

The area of a rectangle, A, is given by A = W * H, so in this case:

32 = W * 2W

32 = 2W²

W² = 16

W = 4

Knowing that the width is 4, the height must be 8. This gives us an area of 32.

8 0

2 years ago

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Violet is baking a mixed berry pie that contains blueberries, cherries, blackberries, and raspberries. She uses three times as m

erastova [34]

Answer:

Blackberries= 30

Rasberries=30

Blueberries=60

Step-by-step explanation:

10 x 3 = 30 blackberries

Black berries is the same amount as rasberries

30 x 2= 60 blueberries

7 0

2 years ago

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For what value of x does 84^3x=512^2x+12

stealth61 [152]

2.193 X 10^-5

this is the answer

7 0

2 years ago

PRECALCULUS QUESTION PLEASE HELP!!!!!!

Shkiper50 [21]

Some information is missing for #6.
#7: use sine, sin35=h/5.1, h=5.1*sin35, use a calculator, h≈2.9
#23: to find the reference angle, keep subtracting the number by 360, until the remaining difference is the smallest positive number.
1406-360-360-360=326. the reference angle is 326.
now look at the remainder, the terminal line of 326 degree is in the 4th quadrant between 270 and 360, the the reference angle is 360-326=34, the angle between the terminal line and the positive x axis. 34 is the answer.

4 0

2 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var

-BARSIC- [3]

Answer:

A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

A)

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}$

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.9

11.2

11.5

11.8

n=10(even)

$Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7$

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}$

Standard deviation :0.89916

C)

8.3

,10.3

,10.5

,10.6

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

,10.3

,10.5

,10.6

,10.7

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5$

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2$

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range : $(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)$

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

3 0

2 years ago