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3 years ago

Whoever helps with number 10 gets brainliest answer and 10 points :)

Mathematics

2 answers:

Irina18 [472]3 years ago

8 0

Before working with any range of numbers you need to put them from smallest to largest, so it would be: (15, 43, 46, 48, 53, 69, 90).

To find the mean you add up all the numbers and divide by how many numbers there are in the sequence. (In this case there are 7).
(15+43+46+48+53+69+90)/7= 52
Mean= 52

To find the median you find the middle most number of the sequence, which is 48. If you need help doing this, all you do is cross out one number at a time at the beginning then at the end, continuously.
Median= 48

To find the mode of this sequence you see which number appears the most often, but in this case there isn't a number that appears more than once.
Mode= no mode

Finding the range is pretty simple, you subtract the largest number by the smallest number (90-15)
Range= 75

Hope this helped!! :)

madreJ [45]3 years ago

3 0

10) Mean 52 Mode No Mode (all shown only once) Median 48 Range 75

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PLEASE ANSWER ASAP REALLY EASY QUESTION WILL GET BRAINLEST!!!!!!!!!!!

-Dominant- [34]

Answer:

random since it doesn't have a specific opinion, it is required by bank service

Step-by-step explanation:

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3 years ago

Can someone help me with this problem

Romashka [77]

Step-by-step explanation:

Find the slope of JK by the formula :y2-y1/(x2-x1) whose value is 4/5

similarly, calculate the slope of GH it's value is also 4/5

Find slope of GK now by same formula y2-y1/( x2-x1 ).the value will be -5/3

Similarly calculate slope of HJ and you will get its value to be -5/3

since parallel lines have equal slopes JK and GH are parallel and GK and HJ are parallel. Since opposite sides are parallel GHJk is parallelogram

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2 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago

2x^2-9x+5 solve from quadratic formula

Black_prince [1.1K]

Answer is either x= 1/2 x=-5

7 0

2 years ago

Read 2 more answers

SOLVE AND SHOW WORK PLSSSS!! Find the length of the missing sides. Do not round (leave in simplified radical form).

Paha777 [63]

Answer:

For the first triangle the missing sides are 7 and 7√2 For the second triangle,the missing sides are 13√3 and 13

Step-by-step explanation:

For the first triangle,

cos(45°)=7/hypotenuse

hypotenuse×cos45=7

hypotenuse×√2/2=7

hypotenuse=14/√2=7√2

We can now find the adjacent side using Pythagorean theorem,

√(7√2)²-7²=7

For the second triangle,

cos(30°)=adjacent/hypotenuse

but cos 30=√3/2

√3/2=adjacent/26(cross multiply and simplify)

26√3/2=13√3

To find the opposite length,

√(26)²-(13√3)²=13

8 0

2 years ago