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katovenus [111]
2 years ago
12

Somebody help me please

Mathematics
2 answers:
Ghella [55]2 years ago
7 0
Answer: m=26

Explanation:

-m/3.25=-8

-m=-8x3.25

-m=-26

m=26
vlabodo [156]2 years ago
5 0
<h3>Solution:</h3>

\frac{m}{ - 3.25}  =  - 8 \\  =  > m =  - 8 \times  - 3.25 \\  =  > m = 26

<h3>Answer:</h3>

C. m = 26

Hope it helps

ray4918 here to help

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I need 3 questions answered please.​
kakasveta [241]

Answer:

3) \:  \: 113 {ft}^{2}

4) \: false

5) \: 70 {ft}^{2}

Step-by-step explanation:

3) Area of a circle is

\pi \times  {r}^{2}

Where r = 6

Thus,

area =   \frac{22}{7}  \times  {6}^{2}

area =  \frac{22}{7}  \times 36

area = 3.143 \times 36

area = 113 {ft}^{2} \: (nearest \: whole \: number)

4) False

Area of a circle is measured by the formula below:

area = \pi \times  {r}^{2}

5) The formula for calculating the area of a rectangle is given thus:

area \: of \: a \: rectangle = l \times w

Where, l = 10ft; w = 7ft

area = 10ft \times 7ft = 70 {ft}^{2}

8 0
3 years ago
The line segment AB is shown below. Which transformation of AB will produce a line segment A’B’ that is parallel to AB
Soloha48 [4]

Answer:

Are there answer choices or do you have to type it in yourself?

4 0
3 years ago
The number of boots that 25 students had in their homes in Florida were recorded below:
Doss [256]

Answer:

A) 8

B) Significantly (increased)

C) Not affected

D) Median

E) Significantly (increased)

F) Not affected

G) Interquartile range

Step-by-step explanation:

A)

When we have a certain set of data, we call "outlier" a value in this set of data that is significantly far from all the other values.

In this problem, we see that we have the following set of data:

0,0,0,0,0,0,0,1,1,1,1,2,2,2,2,2,2,2,3,3,3,3,4,5,8

We see that all the values between 0 and 5 are relatively close to each other, being only 1 number distant from any other value.

On the other hand, the last value "8" is 3 numbers distant from the other highest value ("5"), so the outliar is 8.

B)

The mean of a certain set of value indicates the "average" value of the dataset. It is given by:

\bar x = \frac{\sum x_i}{N}

where

\sum x_i is the sum of all the values in the dataset

N is the number of values in the dataset (in this case, 25)

An outlier usually affects a lot the mean, because it changes the value of the sum \sum x_i by quite a lot.

In fact in this problem, we have

\sum x_i = 47

N = 25

So the mean is

\bar x=\frac{47}{25}=1.88

Instead, without the outliar "8",

\sum x_i = 39\\N=24

And the mean would be

\bar x=\frac{39}{24}=1.63

So we see that the outlier increases the mean significantly.

C)

The median of a dataset is the "central value" of the dataset: it means it is the value for which 50% of the values of the dataset are above the median and 50% of the values of the dataset are below the median.

In this problem, we have 25 values in the dataset: so the median is the 13th values of the ordered dataset (because if we take the 13th value, there are 12 data below it, and 12 data above it). The 13th value here is a "2", so the median is 2.

We see also that the outlier does not affect the median.

In fact, if we had a "5" instead of a "8" (so, not an outlier), the median of this dataset would still be a 2.

D)

The center of a distribution is described by two different quantities depending on the conditions:

- If the distribution is symmetrical and has no tails/outliers, then the mean value is a good value that can be used to describe the center of the distribution.

- On the other hand, if the distribution is not very symmetrical and has tails/outliers, the median is a better estimator of the center of the distribution, because the mean is very affected by the presence of outliers, while the median is not.

So in this case, since the distribution is not symmetrical and has an outlier, it is better to use the median to describe the center of the distribution.

E)

The standard deviation of a distribution is a quantity used to describe the spread of the distribution. It is calculated as:

\sigma = \sqrt{\frac{\sum (x_i-\bar x)^2}{N}}

where

\sum (x_i-\bar x)^2 is the sum of the residuals

N is the number of values in the dataset (in this case, 25)

\bar x is the mean value of the dataset

The presence of an outlier in the distribution affects significantly the standard deviation, because it changes significantly the value of the sum \sum (x_i-\bar x)^2.

In fact, in this problem,

\sum (x_i-\bar x)^2=84.64

So the standard deviation is

\sigma = \sqrt{\frac{84.64}{25}}=1.84

Instead, if we remove the outlier from the dataset,

\sum (x_i-\bar x)^2=45.63

So the standard deviation is

\sigma = \sqrt{\frac{45.63}{24}}=1.38

So, the standard deviation is much higher when the outlier is included.

F)

The interquartile range (IQ) of a dataset is the difference between the 3rd quartile and the 1st quartile:

IQ=Q_3-Q_1

Where:

Q_3 = 3rd quartile is the value of the dataset for which 25% of the values are above Q_3, and 75% of the values are below it

Q_1 = 1st quartile is the value of the dataset for which 75% of the values are above Q_1, and 25% of the values are below it

In this dataset, the 1st and 3rd quartiles are the 7th and the 18th values, respectively (since we have a total of 25 values), so they are:

Q_1=0\\Q_3=2

So the interquartile range is

IQ=2-0 = 2

If the outlier is removed, Q_1, Q_3 do not change, therefore the IQ range does not change.

G)

The spread of a distribution can be described by using two different quantities:

- If the distribution is quite symmetrical and has no tails/outliers, then the standard deviation is a good value that can be used to describe the spread of the distribution.

- On the other hand, if the distribution is not very symmetrical and has tails/outliers, the interquartile range is a better estimator of the spread of the distribution, because the standard deviation is very affected by the presence of outliers, while the IQ range is not is not.

So in this case, since the distribution is not symmetrical and has an outlier, it is better to use the IQ range to describe the spread of the distribution.

5 0
3 years ago
karas jug contains 2 quarts of apple cider.she fills 8 mugs with cider.each mug can hold six three fourths of fluid ounces.how m
bazaltina [42]
So first of all you need to know that in 1 quart there is 31.9995 ounces so since shes says there is 2 quarts lets do the math:)
31.9995 x 2 = 63.999 ounces 
now she says she took 8 mugs and filled them up each with 6 and 3/4 ounces. lets do the math:)
6 and 3/4 x 8 = 30 ounces
now lets subtract that from our 2 quarts or 63.999 ounces 
63.999 - 30 = 33.999 
so we have 33.999 fluid ounces left in the jar:)
hope this helps lol

8 0
3 years ago
I need help with this (don't answer if you don't know)
SashulF [63]

Answer:

Step-by-step explanation:

Here's a step by step tutorial on your calculator on how to do this.  

Hit "stat" then "1: Edit".  If there are numbers there, arrow up to highlight L1, or L2, or wherever there are numbers.  Hit "clear" then "enter" and the numbers will be gone.  In L1, enter the Practice throws values.  Press 3 then enter, then 12 then enter, then 6 then enter, etc. til all of them are in L1's column.  Then arrow over to L2 and do the same with entering all the Free Throw values.

When you're done, hit "stat" again, then arrow over once to "calc" and #4 should say LinReg.  That's a linear regression equation.  If you have a TI 83, just hit enter and you'll get the equation in the form y = mx + b.  If you have a TI 84 or 84+, you'll need to arrow down to the word "calculate" and then you'll see the equation.

One thing...if you have not turned your diagnostics on, you wont be able to see the coefficient of determination (the r-squared value).  To make sure it's on:

Hit 2nd, then 0.  You have opened up the catalog which lists every single thing your calculator can do in alphabetical order.  The hit the button UNDER the MATH button (x to the negative 1) and scroll down until you see "diagnosticsON" and hit enter twice.

If you need to recall the linear regression equation, hit "stat", then "calc" then either enter or calculate and you'll get the linear equation again and an r value and an r-squared value.  The closer that number is to 1, the better the data fits that model.  I got that the equation is

y = 2.352x - .852 with an r-squared value of .844

To get the quadratic regression equation, hit "stat" then "calc" then choose #5, QuadReg.  Repeat the process to calculate your equation.  Mine was

y=.073x^2+1.205x+2.555; r^2=.855

And for the exponential regression, choose ExpReg (mine is under 0.  Yours may be some other number.  Just arrow down til you find it, it's there!) My ExpReg equation was

y=4.229(1.170)^x; r^2=.833

It appears that the r-squared value is the highest in the quadratic regression equation, so that's the best fit.

4 0
3 years ago
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