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2 years ago

NEED ANSWER ASAP PLEASE HELP ME OUT

Mathematics

1 answer:

-Dominant- [34]2 years ago

7 0

Answer:

What exactly do you need help with?

Step-by-step explanation:

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Been stressing out on this can anyone help

HACTEHA [7]

Answer:

43+(x-7)=90

43+(x-7)-43=90-43

(x-7)=46

x-7+7=47+7

x=54

Step-by-step explanation:

4 0

2 years ago

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Find the equation of the line that passes through the point (3,14) and is parallel to y=2/3x -6

Ugo [173]

Y=mx+b
m=slope

paralell ines have same slope
y=2/3x-6
slope is 2/3

the equation of aline that passes through (x1,y1) and has a slope of m is
y-y1=m(x-x1)
given
(3,14) and slope is 2/3
y-14=2/3(x-3)
y-14=2/3x-2
y=2/3x+12

4 0

2 years ago

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Reducing fractions 43/64

Alekssandra [29.7K]

Cannot reduce, it is already in simplest form. You can convert it to a decimal though, 0.671875

8 0

2 years ago

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Simplify the expression -3 (b-7)

Nataliya [291]

Simply distribute the -3 across the b and -7.

-3(b - 7) = (-3 * b) + (-3 * -7) = -3b + 21

Hope this was helpful, if it wasn't just drop me a DM explaining what didn't make sense.

5 0

2 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

<u>e)</u> $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

2 years ago