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2 years ago

12

Plzzz help???????????

2 answers:

uysha [10]2 years ago

6 0

Answer:

Its either C. or D.

Step-by-step explanation:

Lina20 [59]2 years ago

5 0

Answer:

the answer is B................

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I really need some one to solve this please!!!

BartSMP [9]

Answer:x=10/3

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Which of the following shapes have an area of 48 square feet? Select all that apply.

yKpoI14uk [10]

I think 2,3,4 or B,C,D would be correct.

The area of a triangle is (b•h)/2
And the area of a rectangle and parallelogram would just be b•h

1) 8(6) = 48/2 =24

2) 9(10 2/3) = 96/2 =48

3) 19.2(2.5) = 48

4) 16(3) = 48

Hope this is right and hope it helps. Have a good day

4 0

3 years ago

The managers of 21 supermarkets counted the number of cars in their parking lots on the same day. The results are shown in the l

mixer [17]

The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: $(\frac{n+1}{2}) th\ term$

Putting n=21

$(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term$

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

<u>For First Half:</u>

98, 100, 101, 102, <u>108, 109</u>, 111,118, 129, 132

$Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5$

<u>For Second Half:</u>

135, 135, 145, 146, 146, 156, 170 176, 180, 180

$Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151$

Now

<u>Interquartile Range:</u>

$IQR = Q_3-Q_1\\= 151-108.5\\=42.5$

Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

#LearnwithBrianly

7 0

3 years ago

Answer correctly please

velikii [3]

Answer:

The answer would be Image A.

8 0

2 years ago

Condensing Logarithmic Expressions In Exercise, use the properties of logarithms to rewrite the expression as the logarithm of a

yKpoI14uk [10]

Answer:

$4ln [\frac{x^2 (x^3-1)}{x-5}]$

Step-by-step explanation:

For this case we have the following expression:

$4[ln(x^3-1) +2ln(x) -ln(x-5)]$

For this case we can apply the following property:

$a log_c (b) = log_c (b^a)$

And we can rewrite the following expression like this:

$2 ln(x) = ln(x^2)$

And we can rewrite like this our expression:

$4[ln(x^3-1) +ln(x^2) -ln(x-5)]$

Now we can use the following property:

$log_c (a) +log_c (b) = log_c (ab)$

And we got this:

$4[ln(x^3-1)(x^2) -ln(x-5)]$

And now we can apply the following property:

$log_c (a) -log_c (b) = log_c (\frac{a}{b})$

And we got this:

$4ln [\frac{x^2 (x^3-1)}{x-5}]$

And that would be our final answer on this case.

8 0

3 years ago