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3 years ago

I really need help. I'm failing this class guys

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

3 0

17. $2u +5.2 \leq 9.4 + u \\2u - u \leq 9.4 -5.2\\u \leq 4.2$

18. $3(f+2) > 2f + 4\\3f + 6 > 2f + 4 \\3f-2f > 4-6\\f > -2\\$ The answer is I.

19. $4 < 6b -2 \leq 28\\4+2 < 6b \leq 28+2\\6 < 6b \leq 30\\6/6 < b \leq 30/6\\1 < b \leq 5$

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Suppose a baseball player had 229 hits in a season. In the given probability distribution, the random variable x represents the

Andru [333]

Answer:

a) summation of p(x)/n ie. (0.1712+...+0.0051)/6=0.16675

b)1

c).var=summation (x-mean) squared /n ie (0.1712-0.16675)squared +...+(0.0051-0.16675)squared/n=0.027351948

SD =square root of variance =0.16538

Step-by-step explanation:

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2 years ago

THE ANSWER I PUT IS JUST A GUESS, IF IT ISN'T THE ANSWER PLEASE TELL ME WHAT IT IS I BEG YOU

Anna [14]

Answer:

the first one

Step-by-step explanation:

can you help me with my question

6 0

2 years ago

Read 2 more answers

Y = -5 + 6 -----------

andreyandreev [35.5K]

Answer:

y = -5 + 6=1 is your answer

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3 years ago

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A spinner used in a board game is divided into 12 equally sized sectors. Seven of these sectors indicate that the player should

forsale [732]

Answer:

9.3 inches

Step-by-step explanation:

The computation of the radius of the spinner be given below:

Let us assume the same be x

We know that

Total area of the sector = one-sixth of the total area of the spinner

Given that

45.1 = 1 ÷ 6 × π × r^2

r^2 = 45.1 × 6 ÷ π

r^2 = 86.1

r = 9.279

= 9.3 inches

3 0

3 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago